Exercícios sobre álgebra elementar

Assinale a alternativa falsa:

a) $5^{\large 3}\times 5^{\large 4} = 5^{\large 7}$
b) $2^{\large 5} \div 2^{\large 3} = 2^{\large 2}$
c) $(0,5)^{\large 2} \times (0,2)^{\large 2} = (0,1)^{\large 2} = 0,01$
d) $(0,4)^{\large -2} \times (-5)^{\large -2} = 0,25$
e) $\dfrac{(0,05)^{\large 3}}{5^{\large 3}}\,=\,10^{\large -3}$

resposta: (E)
×

Calcular: $\,2^3\,$,$\;(-2)^3\,$ e $\,-2^3$

resposta:

Resolução:

a)	$2^3\,=\,2\centerdot 2 \centerdot 2 \,=\,8$
b)	$(-2)^3 \,=\,(-2)\centerdot(-2)\centerdot (-2) \,=\, -8 $
c)	$-2^3 \,=\, - 2 \centerdot 2 \centerdot 2 \,=\, - 8$

Resposta:

$2^3 \,=\,8\,$, $\; (-2)^3 \,=\, -8\,$ e $\;-2^3 \,=\, -8$

×

Calcular: $\,2^4\,$,$\;(-2)^4\,$ e $\;-2^4\,$.

resposta:

Resolução:

a)	$2^4\,=\,2\centerdot 2 \centerdot 2 \centerdot 2\,=\,16$
b)	$(-2)^4 \,=\,(-2)\centerdot (-2)\centerdot (-2)\centerdot (-2) \,=\, 16 $
c)	$-2^4 \,=\, - 2 \centerdot 2 \centerdot 2 \centerdot 2 \,=\, - 16$

Resposta:

$2^4 \,=\,16\,$, $\;(-2)^4 \,=\, 16\,$, $\;-2^4 \,=\, -16$
×

Calcular: $\,(\frac{1}{3})^3\,$,$\;(0,2)^4\,$ e $\,(0,1)^3$

resposta:

Resolução:

a)	$(\frac{1}{3})^3 \,=\,\frac{1}{3}\centerdot \frac{1}{3}\centerdot \frac{1}{3} \,=\, \frac{1}{27} $
b)	$(0,2)^4\,=\,(0,2)\centerdot (0,2) \centerdot (0,2) \centerdot (0,2)\,=\,0,0016$
c)	$(0,1)^3 \,=\, (0,1) \centerdot (0,1) \centerdot (0,1) \,=\, 0,001$

Resposta:

$(\frac{1}{3})^3 \,=\,\frac{1}{27}\,$, $\;(0,2)^4 \,=\, 0,0016\,$, $\;(0,1)^3 \,=\, 0,001$

×

Fatore:

$\,12a^3b^2\,-\,30a^2b^3\,$

$\,xy\,+\,3x\,+\,4y\,+\,12\,$

$\,6ab\,+\,4b^3\,+\,15a^3\,+\,10a^2b^2\,$

$\,a^2 \,-\, 25\,$

$\,x^2\,-\,1\,$

$\,x^4\,-\,1\,$

$\,144\,-\,81a^2b^2\,$

resposta: a)$\,6a^2b^2(2a-5b)\,$
b) $\,(x+4) \centerdot (y+3)\,$
c) $\,(2b+5a^2) \centerdot (3a+2b^2)\,$
d) $\,(a+5) \centerdot (a-5)\,$
e) $\,(x+1) \centerdot (x-1)\,$
f) $\,(x^2+1) \centerdot (x+1) \centerdot (x-1)\,$
g) $\,9(4+3ab) \centerdot (4-3ab)\,$

×

Calcular: $\;2^{\large -3}\,$, $\;(-2)^{\large -3}\,$, $\;-2^{\large -3}\,$

resposta:

Resolução:

$\,2^{\large -3}\,=\,\dfrac{1}{2^{\large 3}}\,=\,\dfrac{1}{2\centerdot 2 \centerdot 2}\,=\,\dfrac{1}{8}\,=\,0,125\,$

$\,(-2)^{\large -3}\,=\,\dfrac{1}{(-2)^{\large 3}}\,=\,\dfrac{1}{(-2)\centerdot (-2) \centerdot (-2)}\,=\,\dfrac{1}{(-8)}\,=\,-\,\dfrac{1}{8}\,=\,-0,125\,$

$\,-2^{\large -3}\,=\,-\dfrac{1}{2^{\large 3}}\,=\,-\dfrac{1}{2\centerdot 2 \centerdot 2}\,=\,-\dfrac{1}{8}\,=\,-0,125\,$

Resposta:

$\;2^{\large -3}\,=\,0,125\; ; \;(-2)^{\large -3}\,=\,-0,125\; ; \;-2^{\large -3}\,=\,-0,125\;$
×

Calcular: $\;10^{\large -1}\,$, $\;10^{\large -2}\,$, $\;10^{\large -5}\,$

resposta:

Resolução:

$\,10^{\large -1}\,=\,\dfrac{1}{10^{\large 1}}\,=\,\dfrac{1}{10}\,=\,0,1\,$

$\,10^{\large -2}\,=\,\dfrac{1}{10^{\large 2}}\,=\,\dfrac{1}{10\centerdot 10}\,=\,\dfrac{1}{100}\,=\,0,01\,$

$\,10^{\large -5}\,=\,\dfrac{1}{10^{\large 5}}\,=\,\dfrac{1}{10\centerdot 10\centerdot 10\centerdot 10\centerdot 10}\,=\,\dfrac{1}{100\,000}\,=\,0,00001\,$

Resposta:

$\;10^{\large -1}\,=\,0,1\; ; \;10^{\large -2}\,=\,0,01\; ; \; 10^{\large -5}\,=\,0,00001\;$
×

Sendo $\phantom{X}n\phantom{X}$ um número natural ($\;n\,\in\,\mathbb{N}\;$), mostrar que $\;2^{\large n}\,+\,2^{\large n+1}\;=\;3\centerdot 2^{\large n}\;$.

resposta: $\;2^{\large n}\,+\,2^{\large n+1}\;=\;2^{\large n}\,+\,2^{\large n}\centerdot 2\;=\;2^{\large n}\centerdot (1\,+\,2)\;=\;3\centerdot 2^{\large n}\;$
×

Sendo $\phantom{X}n\phantom{X}$ um número natural ($\;n\,\in\,\mathbb{N}\;$), mostre que $\;\dfrac{2^{\large n}\,+\,2^{\large n+1}\,+\,2^{\large n+2}}{2^{\large n+3}\,+\,2^{\large n+4}}\,=\,\dfrac{7}{24}\;$.

resposta:

Resolução:$\;\dfrac{2^{\large n}\,+\,2^{\large n+1}\,+\,2^{\large n+2}}{2^{\large n+3}\,+\,2^{\large n+4}}\,=\,\dfrac{2^{\large n}\,+\,2^{\large n}\centerdot 2\,+\,2^{\large n}\centerdot 2^{\large 2}}{2^{\large n}\centerdot 2^{\large 3}\,+\,2^{\large n}\centerdot 2^{\large 4}}\,$ $=\;\dfrac{2^{\large n}\centerdot(1\,+\,2^{\large 1}\,+\,2^{\large 2})}{2^{\large n}\centerdot(2^{\large 3}\,+\,2^{\large 4})}\,=\,\dfrac{2^{\large n}\centerdot 7}{2^{\large n}\centerdot 24}\,=\,\dfrac{7}{24}\;$.

Calcule as seguintes potências:

$\;1^{\large 4}\;$

$\;0^{\large 3}\;$

$\;5^{\large 3}\;$

$\;(-5)^{\large 3}\;$

$\;-5^{\large 3}\;$

$\;5^{\large 2}\;$

$\;(-5)^{\large 2}\;$

$\;-5^{\large 2}\;$

$\;5^{\large -2}\;$

$\;(-5)^{\large -2}\;$

$\;-5^{\large -2}\;$

$\;5^{\large 0}\;$

$\;(-5)^{\large 0}\;$

$\;-5^{\large 0}\;$

$\;5^{\large 2}\centerdot 5^{\large 3}\;$

$\;\dfrac{5^{\large 3}}{5^{\large 2}}\;$

$\;(0,2)^{\large 3}\centerdot (0,5)^{\large 3}\;$

$\;2^{\large 3^2}\;$

$\;(2^{\large 3})^{\large 2}\;$

$\;(-\,\dfrac{2}{5})^{\large -2}\;$

resposta:

125

-125

-25

$\;\frac{1}{25}\;$

$\;-\,\frac{1}{25}\;$

3125

0,001

512

$\;\frac{25}{4}\;$

Mostre que $\phantom{X}3^{\large 2}\,+\,3^{\large 3}\,\neq\,3^{\,2\,+\,3}\;$

resposta: 9 + 27 ≠ 3⁵ ⇒ 36 ≠ 243;

×

Mostre que $\phantom{X}3^{\large 2}\,+\,4^{\large 2}\,\neq\,(3\,+\,4)^{\large 2}\;$

resposta: 9 + 16 ≠ 7² ⇒ 24 ≠ 49;

×

(UNB) O valor de $\;\left( 5^{-5} \right)^{\large 5}\;$ é:

$\;5^{-25}\;$

$\;-\,\dfrac{1}{125}\;$

$\;(-25)^{\large 5}\;$

$\;5^{\large -5}\;$

nenhuma dessas

resposta: (A)
×

(UEMT) Simplificando-se a expressão $\phantom{X}\left[ 2^{\large 9}\,\div \, ( 2^{\large 2} \centerdot 2)^{\large 3}\right]^{\large -3}\phantom{X}$, obtém-se:

$\,2^{\large 36}\,$

$\,2^{\large -30}\,$

$\,2^{\large -6}\,$

$\,1\,$

$\,\dfrac{1}{3}\,$

resposta: (D)
×

(UFSM) Efetuando a divisão $\phantom{X}e^{\large x}\, \div \,e^{\large x\,-\,2}\phantom{X}$, teremos:

$\,e^{\large -2}\,$

$\,e^{\large x^2\,-\,2x}\,$

$\,e^{\large 2}\,$

$\,(e)^{\large \frac{x}{x\,-\,2}}\,$

nenhuma das anteriores

resposta: (C)
×

(FUVEST) Calcule:

a)$\;\dfrac{1}{10}\;-\;\dfrac{1}{6}\;$

b)$\;\dfrac{0,2\; \centerdot\; 0,3}{3,2 \; \centerdot \; 2,0}\;$

resposta: a)-1/15 b)0,05

a)$\;\dfrac{3\,-\,5}{30}\,=\,\dfrac{-2}{30}\,=\;-\,\dfrac{1}{15}\;$
b)$\;\dfrac{0,06}{1,2}\,=\,\dfrac{6 \centerdot 10^{\large -2}}{12\centerdot 10^{\large -1}}\,=\;\dfrac{1}{20}\,=\,0,05\;$

(LONDRINA) Dados os números $\phantom{X}x\,=\,\dfrac{\dfrac{1}{3}\,+\,\dfrac{1}{3}}{\dfrac{1}{3}}\phantom{X}$, $\phantom{X}y\,=\,\dfrac{\dfrac{1}{3}\,+\,\dfrac{1}{3}}{\dfrac{3}{2}}\phantom{X}$, $\phantom{X}z\,=\,\dfrac{\dfrac{\dfrac{1}{3}\,+\,\dfrac{1}{3}}{3}}{\dfrac{1}{2}}\phantom{X}$ pode-se concluir que:

x, y e z são iguais

x > y e y = z

x < y e y = z

x > y e y > z

x > y e (y + z) é inteiro

resposta: (B)
×

(ENG ARARAQUARA) Calcular a expressão:
$\phantom{X}\dfrac{3\,+\,\dfrac{5}{16}\,-\,4\,+\,\dfrac{3}{4}\,-\,\dfrac{1}{2}}{0,0001}\,\centerdot\,0,005\phantom{X}$

resposta: -175/8
×

(FUVEST) O valor da expressão $\phantom{X}\dfrac{a\,+\,b}{1\,-\,ab}\phantom{X}$, para $\;a\,=\,\dfrac{1}{2}\phantom{X}$ e $\phantom{X}b\,=\,\dfrac{1}{3}\;$ é:

resposta: (B)
×

(MACKENZIE) O valor numérico de $\phantom{X}\dfrac{xy\,-\,x^{\large 2}}{\sqrt{y}}\phantom{X}$ para $\,x\,=\,-0,1\;$ e $\;y\,=\,0,01\;$ é:

-0,11

-0,011

-0,0011

0,011

0,11

resposta: (A)
×

(UNB) A expressão $\phantom{X}\dfrac{1\,+\,\dfrac{1}{1\,-\,\dfrac{1}{5}}}{-1\,+\,\dfrac{3}{1\,+\,\dfrac{1}{5}}}\phantom{X}$ é equivalente a:

$\,\dfrac{3}{2}\,$

$\,\dfrac{2}{3}\,$

$\,\dfrac{1}{3}\,$

$\,\dfrac{1}{4}\,$

$\,\dfrac{1}{2}\,$

resposta: (A)
×

(FUVEST) Calcule o valor numérico de $\phantom{X}\dfrac{-x^{\large 2}\,+\,xy}{y}\phantom{X}$ para $\,x\,=\,-0,1\;$ e $\;y\,=\,0,001\;$

resposta: -10,1
×

(UBERABA) O valor de $\phantom{X}ab^{\large 2}\,-\,a^{\large 3}\phantom{X}$ para $\phantom{X}a\,=\,-\,\dfrac{x}{2}\phantom{X}$ e $\phantom{X}b\,=\,2x\phantom{X}$ é:

$\,\dfrac{17}{8}x^{\large 3}\,$

$\,-\,\dfrac{17}{8}x^{\large 3}\,$

$\,-\,\dfrac{15}{8}x^{\large 3}\,$

$\,-\,\dfrac{11}{6}x^{\large 3}\,$

$\,-\dfrac{13}{6}x^{\large 3}\,$

resposta: (C)
×

(SANTA CASA) O valor de $\phantom{X}\dfrac{3^{\large -1}\,+\,5^{\large -1}}{2^{\large -1}}\phantom{X}$ é:

$\,\dfrac{4}{15}\,$

$\,\dfrac{1}{2}\,$

$\,\dfrac{1}{8}\,$

$\,\dfrac{16}{15}\,$

$\,4\,$

resposta: (D)
×

(UFRN) Se $\phantom{X}a\,=\,0,1\phantom{X}$ e $\phantom{X}b\,=\,0,2\phantom{X}$, o valor da expressão $\phantom{X}\dfrac{a^{\large 2}b^{\large 2}\,-\,a^{\large 3}b}{b^{\large 2}\,-\,a^{\large 2}}\phantom{X}$ é:

$\,\dfrac{1}{300}\,$

$\,\dfrac{1}{150}\,$

$\,\dfrac{1}{100}\,$

$\,\dfrac{1}{75}\,$

$\,\dfrac{1}{200}\,$

resposta: (B)
×

Sendo $\phantom{X}x\,=\,(2^{\large 2})^{\large 3}\;$, $\phantom{X}y\,=\,2^{\large 2^3}\phantom{X}$ e $\phantom{X}z\,=\,2^{\large 3^2}\;$, escrevendo o produto $\phantom{X}x\,\centerdot\,y\,\centerdot z\phantom{X}$ na forma $\;2^{\large n}\;$, qual o valor de $\,n\,$?

resposta: 23
×

(OSEC) Sabendo-se que $\phantom{X}a^{\large 2}\,=\,5^{\large 6}\;$, $\phantom{X}b^{\large 3}\,=\,5^{\large 7}\;$, $\phantom{X}c^{\large 4}\,=\,5^{\large 8}\phantom{X}$ e que $\,a\;$ e $\,c\;$ são dois números reais de mesmo sinal, ao escrever $\phantom{X}(a\,b\,c)^{\large 9}\phantom{X}$ como potência de base 5, qual o valor do expoente?

resposta: 66
×

Assinalar a falsa:

Se $\,x^{\large 2}\,=\,4\,$ então $\,x^{\large 6}\,=\,64\,$

Se $\,x^{\large 6}\,=\,64\,$ então $\,x\,=\,2\,$

$\,\left(2^{\large 2}\right)^{\large 3}\,< \,2^{\large 2^3}\,$

Se $\,10^{\large x}\,=\,0,2\,$ então $\,10^{\large 2x}\,=\,0,4\,$

$\,2^{\large n+2}\,+\,2^{\large n}\,= \,5\centerdot\,2^{\large n}\,$

resposta: (B)
×

(OSEC) Se $\phantom{X}10^{\large 2x}\,=\,25\phantom{X}$ então $\,10^{\large -x}\,$ é igual a:

$\frac{1}{5}\,$

$\frac{1}{25}\,$

-5

resposta: (B)
×

(MED SANTO ANDRÉ) Simplificando a expressão $\phantom{X}\dfrac{2^{\large n+4}\,-\,2\,\centerdot \,2^{\large n}}{2\, \centerdot\, 2^{\large n+3}}\phantom{X}$ obtém-se:

$\,2^{\large n+1} - \dfrac{1}{8}\,$

$\,\dfrac{7}{8}\,$

$\,-\,2^{\large n+1}\,$

$\,1\,-\,2^{\large n}\,$

$\,\dfrac{7}{4}\,$

resposta: (B)
×

(PUC DF) Assinale a alternativa correta relativa às afirmativas I. até IV.:

$\,(-3)^{\large -2}\,=\,-\,9\,$

II.

$\,(\dfrac{1}{2})^{\large -1}\,=\,2\,$

III.

$\,(-2)^{\large -3}\,=\,-\,8\,$

IV.

$\,2^{\large -3}\,=\,\dfrac{1}{8}\,$

II e IV estão corretas;

I e III estão corretas;

III e IV são falsas;

III está correta;

Todas são falsas.

resposta: (A)
×

(OBJETIVO - 1982) Simplificando-se a expressão $\phantom{X}\left(2^{\large 3}\right)^{\large 2^3}\phantom{X}$ obtém-se:

$\,6^{\large 6}\,$

$\,6^{\large 8}\,$

$\,2^{\large 8}\,$

$\,2^{\large 18}\,$

$\,2^{\large 24}\,$

resposta: (E)
×

(MED JUNDIAÍ - 1982) Sejam as sentenças abaixo, onde a é um número real tal que 0 < a < 1 .

$\;a^{\large -x}\,=\,-\,a^{\large x}, \,\forall\,x\,\in\;\mathbb{R}$

II.

$\;2a^{\large x}\,+\,\,a^{\large x}\,=\,3\,\centerdot\,a^{\large x}, \,\forall\,x\,\in\;\mathbb{R}$

III.

$\;a^{\large x}\,\centerdot\,a^{\large x}\,=\,\,a^{\large x^2}, \,\forall\,x\,\in\;\mathbb{R}$

apenas I é falsa

apenas II é falsa

apenas III é falsa

apenas I e II são falsas

apenas I e III são falsas

resposta: (E)
×

(CESGRANRIO) A representação decimal de $\;0,01^{\large 3}\;$ é:

0,03

0,001

0,0001

0,000001

0,0000001

resposta: (D)
×

(VUNESP - 1982) Se $\;x\,=\,10^{\large 3}\;$ então
$\phantom{X}\dfrac{(0,1)\,\centerdot\,(0,001)\,\centerdot\,10^{\large -1}}{10\,\centerdot\,(0,0001)}\phantom{X}$
é igual a:

100x

10x

x/10

x/100

resposta: (B)
×

Simplificar $\;\sqrt{48}\;$.

resposta:

Resolução:
$\;\sqrt{48}\;=\,\sqrt{16\,\centerdot\,3}\,=$ $\,\sqrt{16}\,\centerdot\,\sqrt{3}\,=\,4\,\centerdot\,\sqrt{3}$
Resposta:

$\;4\sqrt{3}\;$
×

Calcule:

$\,\sqrt{81}\,$

$\,-\,\sqrt{81}\,$

$\,\sqrt[\large 3]{64}\,$

$\,\sqrt[\large 3]{\,-\,64}\,$

$\,\sqrt{2}\,\centerdot\,\sqrt{50}\,$

$\,\dfrac{\sqrt{27}}{\sqrt{3}}\,$

$\,\sqrt[\large 3]{\sqrt{64}}\,$

resposta: a.9; b.-9; c.4; d.-4; e.10; f.3; g.2;
×

Mostre que $\phantom{X}\sqrt{9\,+\,16}\,\neq\,\sqrt{9}\,+\,\sqrt{16}\phantom{X}$

resposta: $\,\sqrt{9\,+\,16}\,\neq\,\sqrt{9}\,+\,\sqrt{16}\,\Rightarrow$ $\,\sqrt{25}\,\neq\,3\,+\,4\,\Rightarrow\,5\,\neq\,7\,$c.q.d.
×

(PUC DF) Assinale a correta:

$\,\sqrt[\large 3]{-27}\,=\,-\,3\,$

II.

$\,5^{-\,\frac{1}{2}}\,=\,5\,$

III.

$\,\dfrac{1}{\sqrt{3}}\,=\,\dfrac{\sqrt{3}}{3}\,$

IV.

$\,\sqrt[\large 3]{2^{\large 5}}\,=\,2^{\large \frac{3}{5}}\,$

II e III estão corretas

I e IV estão corretas

I e III estão corretas

todas estão corretas

todas estão erradas

resposta: (C)
×

O valor numérico da expressão $\phantom{X} {2}\,\sqrt{xy^{\phantom{X}}}\,-\,\sqrt{x^{\large 2}\,-\,21y}\phantom{X}$, para x = 12 e y = 3 , é igual a:

-3

resposta: (D)
×

Calcular o seguinte:

$\phantom{X}\left(\sqrt{9\,\times\,9}\right)^{\large 2}\phantom{X}$

$\phantom{X}\left(\sqrt{3\,\times\,27}\right)^{\large 2}\phantom{X}$

$\phantom{X}\left(\sqrt{6^{\large 2}}\right)^{\large 2}\phantom{X}$

$\phantom{X}\dfrac{\sqrt{27\,-\,27}}{\sqrt{81}}\phantom{X}$

$\phantom{X}\sqrt{2\,\times\,2}\,+\,\sqrt{49}\phantom{X}$

$\phantom{X}\sqrt{7\,\times\,7}\,-\,\sqrt{100}\phantom{X}$

$\phantom{X}\dfrac{\sqrt{20^{\large 2}}}{\sqrt{81}}\phantom{X}$

$\phantom{X}\left(\sqrt{5^{\large 2}}\right)^{\large 2}\phantom{X}$

$\phantom{X}\dfrac{\sqrt{0\,+\,0}}{\sqrt{25}}\phantom{X}$

$\phantom{X}\sqrt{49}\,\times\,\sqrt{140\,-\,76}\phantom{X}$

resposta: a.81 b.81 c.36 d.0 e.9 f.-3 g.20/9 h.25 i.0 j.56
×

Calcule as raízes seguintes com aproximação de 1 casa decimal.

$\phantom{X}\sqrt{2,89}\phantom{X}$

$\phantom{X}\sqrt{1,44}\phantom{X}$

$\phantom{X}\sqrt{0,09}\phantom{X}$

$\phantom{X}\sqrt{2,56}\phantom{X}$

$\phantom{X}\sqrt{1,69}\phantom{X}$

$\phantom{X}\sqrt{0,16}\phantom{X}$

$\phantom{X}\sqrt{3,24}\phantom{X}$

$\phantom{X}\sqrt{0,49}\phantom{X}$

$\phantom{X}\sqrt{3,61}\phantom{X}$

$\phantom{X}\sqrt{0,04}\phantom{X}$

$\phantom{X}\sqrt{0,81}\phantom{X}$

$\phantom{X}\sqrt{0,25}\phantom{X}$

$\phantom{X}\sqrt{0,36}\phantom{X}$

$\phantom{X}\sqrt{0,01}\phantom{X}$

$\phantom{X}\sqrt{4,41}\phantom{X}$

$\phantom{X}\sqrt{2,25}\phantom{X}$

resposta: a.1,7 b.1,2 c.0,3 d.1,6 e.1,3 f.0,4 g.1,8 h.0,7 i.1,9 j.0,2 k.0,9 l.0,5 m.0,6 n.0,1 o.2,1 p.1,5
×

(UBERLÂNDIA) Qual a afirmativa certa?

$\phantom{X}\sqrt{25\,+\,16}\,=\,9\phantom{X}$

$\phantom{X}\sqrt{5}\,+\,\sqrt{5}\,=\,\sqrt{10}\phantom{X}$

$\phantom{X}5\sqrt{2}\,<\,\sqrt{20}\phantom{X}$

$\phantom{X}0,2\,<\,\sqrt{4}\phantom{X}$

$\phantom{X}3\sqrt{10}\,=\,\sqrt{30}\phantom{X}$

resposta: (D)
×

(UEMT) O número $\phantom{X}\sqrt{2352}\phantom{X}$ corresponde a:

$\,4\,\sqrt{7}\,$

$\,4\,\sqrt{21}\,$

$\,28\,\sqrt{3}\,$

$\,28\,\sqrt{21}\,$

$\,56\,\sqrt{3}\,$

resposta: (C)
×

(UNB) A expressão $\phantom{X}\left({\large 2}\,^{1/2}\right)^{-\,1/2}\phantom{X}$ equivale a:

$\phantom{X}\sqrt{2}\phantom{X}$

$\phantom{X}\sqrt[4]{2}\phantom{X}$

$\phantom{X}\sqrt{\dfrac{1}{\sqrt{2}}}\phantom{X}$

$\phantom{X}\dfrac{1}{\sqrt{2}}\phantom{X}$

nenhuma das anteriores

resposta: (D)
×

(FGV) $\phantom{X}{\large \dfrac{\,2\,}{3}\,\centerdot\,\left(8\right)^{\frac{\,2\,}{3}}\,-\,\dfrac{\,2\,}{3}\,\centerdot\,\left(8\right)^{-\,\frac{\,2\,}{3}}}\phantom{X}$ é igual a:

-1

2,5

0,5

resposta: (C)
×

Simplifique as seguintes raízes:

$\phantom{X}\sqrt{180}\phantom{X}$

$\phantom{X}\sqrt{129}\phantom{X}$

$\phantom{X}\sqrt{135}\phantom{X}$

$\phantom{X}\sqrt{48}\phantom{X}$

$\phantom{X}\sqrt{155}\phantom{X}$

$\phantom{X}\sqrt{31}\phantom{X}$

$\phantom{X}\sqrt{6}\phantom{X}$

$\phantom{X}\sqrt{21}\phantom{X}$

$\phantom{X}\sqrt{50}\phantom{X}$

$\phantom{X}\sqrt{24}\phantom{X}$

$\phantom{X}\sqrt{43}\phantom{X}$

$\phantom{X}\sqrt{9}\phantom{X}$

$\phantom{X}\sqrt{89}\phantom{X}$

$\phantom{X}\sqrt{114}\phantom{X}$

$\phantom{X}\sqrt{26}\phantom{X}$

$\phantom{X}\sqrt{19}\phantom{X}$

resposta: a. $\,6\sqrt{5}\,$ b. $\,\sqrt{129}\,$ c. $\,3\sqrt{15}\,$ d. $\,4\sqrt{3}\,$ e. $\,\sqrt{155}\,$ f. $\,\sqrt{31}\,$ g. $\,\sqrt{6}\,$ h. $\,\sqrt{21}\,$ i. $\,5\sqrt{2}\,$ j. $\,2\sqrt{6}\,$ k. $\,\sqrt{43}\,$ l. $\,3\,$ m. $\,\sqrt{80}\,$ n. $\,\sqrt{114}\,$ o. $\,\sqrt{26}\,$ p. $\,\sqrt{19}\,$
×

(OSEC) Escolha a alternativa correta:

$\phantom{X}2\sqrt{3}\,+\,5\sqrt{3}\,=\,7\sqrt{3}\,\centerdot\,\sqrt{3}\,=$ $\,21\,+\,\dfrac{\,1\,}{2}\,=\,\dfrac{\,43\,}{2}\phantom{X}$

$\phantom{X}5\sqrt{2}\,+\,6\sqrt{3}\,=\,11\sqrt{5}\phantom{X}$

$\phantom{X}18\sqrt{3}\,-\,6\sqrt{3}\,+\,4\sqrt{3}\,=\,8\sqrt{3}\phantom{X}$

$\phantom{X}4\sqrt{3}\,+\,2\sqrt{3}\,\centerdot\,5\sqrt{3}\,=\,90\phantom{X}$

nenhuma das alternativa anteriores é correta

resposta: (E)
×

Fatorar:$\phantom{X}6a^{\large 2}b\,+\,8a\phantom{X}$

resposta:

Resolução:$\,6a^{\large 2}b\,+\,8a\,=\,2a(3ab\,+\,4)\,$

$\,2a(3ab\,+\,4)\,$
×

Fatorar:$\phantom{X}a^{\large 4}b^{\large 3}c^{\large 3}\,+\,a^{\large 3}b^{\large 4}c^{\large 3}\,+\,a^{\large 3}b^{\large 3}c^{\large 4}\phantom{X}$

resposta:

Resolução:$\,a^{\large 4}b^{\large 3}c^{\large 3}\,+\,a^{\large 3}b^{\large 4}c^{\large 3}\,+\,a^{\large 3}b^{\large 3}c^{\large 4}\,=$ $\,a^{\large 3}ab^{\large 3}c^{\large 3}\,+\,a^{\large 3}b^{\large 3}bc^{\large 3}\,+\,a^{\large 3}b^{\large 3}c^{\large 3}c\,=$

$\,a^{\large 3}b^{\large 3}c^{\large 3}(a\,+\,b\,+\,c)\,$
×

Fatorar:$\phantom{X}x^{\large 6}\,-\,5x^{\large 5}\,+\,26x^{\large 4}\phantom{X}$

resposta:

Resolução:$\,x^{\large 6}\,-\,5x^{\large 5}\,+\,26x^{\large 4}\,=$

$\,x^{\large 4}\,\centerdot\,(x^{\large 2}\,-\,5x\,+\,26)\,$
×

Fatorar:$\phantom{X}a^{\large 4}\,-\,a^{\large 3}\phantom{X}$

resposta:

Resolução:$\,a^{\large 4}\,-\,a^{\large 3}\,=$

$\,a^{\large 3}\,\centerdot\,(a\,-\,1)\,$
×

Desenvolva:

$\phantom{X}(2\,+\,3m)^{\large 2}\phantom{X}$

$\phantom{X}(a\,-\,3)^{\large 2}\phantom{X}$

$\phantom{X}(\sqrt{5}\,+\,\sqrt{3})^{\large 2}\phantom{X}$

$\phantom{X}(a\,+\,3b)^{\large 3}\phantom{X}$

$\phantom{X}(2a\,-\,b)^{\large 3}\phantom{X}$

resposta: a. $\,4\,+\,12m\,+\,9m^{\large 2}\,$ b. $\,a^{\large 2}\,-\,6a\,+\,9\,$ c. $\,5\,+\,2\sqrt{15}\,+\,3\,=\,8\,+\,2\sqrt{15}\,$ d. $\,a^{\large 3}\,+\,9a^{\large 2}b\,+\,27ab^{\large 2}\,+\,27b^{\large 3}\,$ e. $\,8a^{\large 3}\,+\,12a^{\large 2}b\,+\,6ab^{\large 2}\,-b^{\large 3}\,$
×

Fatore:

a.$\phantom{X}1\,+\,6a\,+\,12a^{\large 2}\,+\,8a^{\large 3}\phantom{X}$
b.$\phantom{X}x^{\large 3}\,-\,6x^{\large 2}y\,+\,12xy^{\large2}\,-\,8y^{\large 3}\phantom{X}$

resposta: a. (1 + 2a)³ b. (x - 2y)³
×

Fatore:

$\phantom{X}a^{\large 2}\,+\,4a\,+\,4\phantom{X}$

$\phantom{X}9a^{\large 2}\,+\,30ab\,+\,25b^{\large 2}\phantom{X}$

$\phantom{X}1\,-\,18x^{\large 2}\,+\,81x^{\large 4}\phantom{X}$

$\phantom{X}a^{\large 3}\,+\,27\phantom{X}$

$\phantom{X}64\,-\,x^{\large 3}\phantom{X}$

resposta: a. (a + 2)² b. (3a + 5b)² c. (1 - 9x²)² d. (a + 3)(a² - 3a + 9) e. (4 - x)(x² + 4x + 16)
×

Fatorar:$\phantom{X}(a\,+\,b)\,\centerdot\,x\,+\,2(a\,+\,b)\phantom{X}$

resposta: $(a\,+\,b)(x\,+\,2)$
×

Fatorar: $\phantom{X}2x\,+\,ax\,+\,2y\,+\,ay\phantom{X}$

resposta: Resolução:
$\rlap{ \underbrace{\phantom{2x\,+\,ax}}_{\small x(2\,+\,a)} }2x\,+\,ax +\underbrace{2y\,+\,ay}_{\small y(2\,+\,a)} \,= $ $\,x\centerdot(2\,+\,a)\,+\,y\centerdot(2\,+\,a)\,=\,$ (2 + a)(x + y)
×

Fatorar: $\phantom{X}x^{{}^{\Large 3}}\,+\,x^{{}^{\Large 2}}\,-\,3x\,-3\phantom{X}$

resposta: Resolução:
$\rlap{ \underbrace{\phantom{x^{{}^{\Large 3}}\,+\,x^{{}^{\Large 2}}\,}}_{\small x^{{}^{2}}(x\,+\,1)} }x^{{}^{\Large 3}}\,+\,x^{{}^{\Large 2}}\, - \,\underbrace{3x\,-\,3}_{\small -3(x\,+\,1)} \,= $ $\,x^{{}^{\large 2}}\centerdot(x\,+\,1)\,-\,3\centerdot(x\,+\,1)\,=\,$ (x + 1)(x² - 3)
×

Fatorar: $\phantom{X}x^{{}^{\Large 3}}\,+\,x^{{}^{\Large 2}}\,+\,x\,+\,1\phantom{X}$

resposta: Resolução:
$\rlap{ \underbrace{\phantom{x^{{}^{\Large 3}}\,+\,x^{{}^{\Large 2}}\,}}_{\small x^{{}^{2}}(x\,+\,1)} }x^{{}^{\Large 3}}\,+\,x^{{}^{\Large 2}}\, + \,\underbrace{x\,+\,1}_{\small 1(x\,+\,1)} \,= $ $\,x^{{}^{\large 2}}\centerdot(x\,+\,1)\,+\,1\centerdot(x\,+\,1)\,=\,$ (x + 1)(x² + 1)
×

Fatorar: $\phantom{X}x^{{}^{\Large 3}}\,-\,x^{{}^{\Large 2}}\,-\,x\,+\,1\phantom{X}$

resposta: Resolução:
$\rlap{ \underbrace{\phantom{x^{{}^{\Large 3}}\,-\,x^{{}^{\Large 2}}\,}}_{\small x^{{}^{2}}(x\,-\,1)} }x^{{}^{\Large 3}}\,-\,x^{{}^{\Large 2}}\, - \,\underbrace{x\,+\,1}_{\small -1(x\,-\,1)} \,= $ $\,x^{{}^{\large 2}}\centerdot(x\,-\,1)\,-\,1\centerdot(x\,-\,1)\,=\,$ (x - 1)(x² - 1)
×

Fatorar: $\phantom{X}x^{{}^{\Large 2}}\,-\,5x\,+\,6\phantom{X}$

resposta: Resolução:
$\,x^{{}^{\Large 2}}\,-\,5x\,+\,6\,=\,\rlap{ \underbrace{\phantom{x^{{}^{\Large 2}}\,-\,2x\,}}_{\small x(x\,-\,2)} }x^{{}^{\Large 2}}\,-\,2x\, - \,\underbrace{3x\,+\,6}_{\small -3(x\,-\,2)} \,= $ $\,x(x\,-\,2)\,-\,3\centerdot(x\,-\,2)\,=\,$ (x - 3)(x - 2)
×

Fatorar: $\phantom{X}x^{{}^{\Large 2}}\,+\,2y^{{}^{\Large 2}}\,+\,3xy\,+\,x\,+\,y\phantom{X}$

resposta: Resolução:
$\,x^{{}^{\Large 2}}\,+\,2y^{{}^{\Large 2}}\,+\,3xy\,+\,x\,+\,y\,=$ $\,x^{{}^{\Large 2}}\,+\,2y^{{}^{\Large 2}}\,+\,xy\,+\,2xy\,+\,x\,+\,y\,=$ $\,\,x^{{}^{\Large 2}}\,+\,xy\,+\,2y^{{}^{\Large 2}}\,+\,2xy\,+\,x\,+\,y\,=$ $\,x(x\,+\,y)\,+\,2y\centerdot(x\,+\,y)\,+\,1\centerdot(x\,+\,y)\,=$ (x + y)(x + 2y + 1)
×

Fatorar: $\phantom{X}4a^{{}^{\Large 2}}\,-\,9b^{{}^{\Large 2}}\phantom{X}$

resposta: Resolução:
$\,4a^{{}^{\Large 2}}\,-\,9b^{{}^{\Large 2}}\,=$ $\,2^{{}^{\Large 2}}\,\centerdot\,a^{{}^{\Large 2}}\,-\,3^{{}^{\Large 2}}\,\centerdot \,b^{{}^{\Large 2}}\,=$ $\,(2a)^{{}^{\Large 2}}\,-\,(3b)^{{}^{\Large 2}}\,=$ $\,(2a\,+\,3b)\,\centerdot\,(2a\,-\,3b)\phantom{X}=$ (2a + 3b)(2a - 3b)
×

Fatorar: $\phantom{X}(x\,+\,y)^{{}^{\Large 2}}\,-\,y^{{}^{\Large 2}}\phantom{X}$

resposta: Resolução:
$\,(x\,+\,y)^{{}^{\Large 2}}\,-\,y^{{}^{\Large 2}}\,=$ $\,[(x\,+\,y)\,+\,y\,]\centerdot[(x\,+\,y)\,-\,y\,]\,=$ $\,(x\,+\,2y)\centerdot x\,=\phantom{X}$ x(x + 2y)
×

Fatorar: $\phantom{X}(x\,+\,y)^{{}^{\Large 2}}\,-\,(x\,-\,y)^{{}^{\Large 2}}\phantom{X}$

resposta: Resolução:
$\,(x\,+\,y)^{{}^{\Large 2}}\,-\,(x\,-\,y)^{{}^{\Large 2}}\,=$ $\,[(x\,+\,y)\,+\,(x\,-\,y)\,]\centerdot[(x\,+\,y)\,-\,(x\,-\,y)\,]\,=$ $\,[x\,+\,y\,+\,x\,-\,y\,]\centerdot[x\,+\,y\,-\,x\,+\,y\,]\,=$ $\,2x\,\centerdot\,2y\,=\phantom{X}$ 4xy
×

Fatorar: $\phantom{X}1\,-\,(x\,+\,y)^{{}^{\Large 2}}\phantom{X}$

resposta: Resolução:
$\,1\,-\,(x\,+\,y)^{{}^{\Large 2}}\,=$ $\,1^{{}^{\Large 2}}\,-\,(x\,+\,y)^{{}^{\Large 2}}\,=$ $\,[1\,+\,(x\,+\,y)\,]\centerdot[1\,-\,(x\,+\,y)\,]\,=$ $\,(1\,+\,x\,+\,y)\,\centerdot\,(1\,-\,x\,-\,y)\,=\phantom{X}$ (1 + x + y)(1 - x - y)
×

Fatorar: $\phantom{X}x^{{}^{\Large 4}}\,-\,y^{{}^{\Large 4}}\phantom{X}$

resposta: Resolução:
$\,x^{{}^{\Large 4}}\,-\,y^{{}^{\Large 4}}\,=$ $\,(x^{{}^{\Large 2}})^{{}^{\Large 2}}\,-\,(y^{{}^{\Large 2}})^{{}^{\Large 2}}\,=$ $\,(x^{{}^{\Large 2}}\,+\,y^{{}^{\Large 2}})\centerdot(x^{{}^{\Large 2}}\,-\,y^{{}^{\Large 2}})\,$
$\,(x^{{}^{\Large 2}}\,+\,y^{{}^{\Large 2}})\centerdot(x^{{}^{\Large 2}}\,-\,y^{{}^{\Large 2}})\,$ é uma expressão fatorada (e portanto a resposta do exercício). Mas (x² - y²) é uma diferença de quadrados, então podemos continuar:
$\,(x^{{}^{\Large 2}}\,+\,y^{{}^{\Large 2}})\centerdot(x^{{}^{\Large 2}}\,-\,y^{{}^{\Large 2}})\,=$ $(x^{{}^{\Large 2}}\,+\,y^{{}^{\Large 2}})\centerdot(x\,+\,y)\centerdot(x\,-\,y)\;=\;$ (x² + y²)(x + y)(x - y)
×

Fatorar: $\phantom{X}a^2\,+\,6a\,+\,9\phantom{X}$

resposta: a² + 6a + 9 = a² + 2 . a . 3 + 3² = (a + 3)²
×

Fatorar: $\phantom{X}25x^2\,+\,70x\,+\,49\phantom{X}$

resposta: 25x² + 70x + 49 = (5x)² + 2 . 5x . 7 + 7² = (5x + 7)²
×

Fatorar: $\phantom{X}x^2\,-\,2x\,+\,1\phantom{X}$

resposta: x² - 2x + 1 = (x)² - 2 . x . 1 + 1² = (x - 1)²
×

Fatorar: $\phantom{X}a^3\,-\,10a^2\,+\,25a\phantom{X}$

resposta: a³ - 10a² + 25a = a.[a² - 10a + 25] = a.[(a)² - 2 . a . 5 + 5² = a(a - 5)²
×

Sabendo que $\phantom{X}a\,+\,\dfrac{1}{a}\,=\,3\phantom{X}$, calcular o valor de $\phantom{X}a^2\,+\,\dfrac{1}{a^2}\phantom{X}$

resposta: Resolução:
$\,a\,+\,\dfrac{1}{a}\,=\,3\;\Rightarrow$ $\;\left(a\,+\,\dfrac{1}{a}\right)^2\,=\,9\;\Leftrightarrow$ $\;a^2\,+\,2\,\centerdot\,a\,\centerdot\,\dfrac{1}{a}\,+\,\dfrac{1}{a^2}\,=\,9\;\Leftrightarrow$ $\;a^2\,+\,2\,+\,\dfrac{1}{a^2}\,=\,9\;\Leftrightarrow$ $\;a^2\,+\,\dfrac{1}{a^2}\,=\,9\,-\,2\,=$ $\,7$
×

Simplificar as expressões abaixo, admitindo que todos os denominadores são diferentes de zero.

$\;\dfrac{\;x^2\,+\,2xy\,+\,y^2}{x^2\,-\,y^2}\;$

$\;\dfrac{a^3\,-\,1}{a^2\,-\,1}\;$

$\;\dfrac{m^3\,+\,n^3}{m^3\,-\,m^2n\,+\,mn^2}\;$

$\;\dfrac{x^3\,+\,3x^2y\,+\,3xy^2\,+\,y^3}{x^3\,+\,y^3}\,\div\,\dfrac{x^2\,+\,2xy\,+\,y^2}{x^2\,-\,xy\,+\,y^2}\;$

resposta:

$\,\frac{x+y}{x-y}\,$

$\,\frac{a^2+a+1}{a+1}\,$

$\,\frac{m+n}{m}$

Fatorar $\phantom{X}x^6\,+x^2y^4\,-\,x^4y^2\,-\,y^6\phantom{X}$

resposta: $\,(x^4\,+\,y^4)(x\,+\,y)(x\,-\,y)\;$
×

Fatorar $\phantom{X}a^2\,+\,b^2\,-\,c^2\,-\,2ab\phantom{X}$

resposta: (a - b - c).(a - b + c)
×

(MED SANTOS) Calcular $\phantom{X}934287^2 - 934286^2\;$

1868573

1975441

nenhuma das anteriores

resposta: (A)
×

(ENG ARARAQUARA) Simplificar $\phantom{X}\dfrac{\;a^{{}^{\Large -4}}\,-\,b^{{}^{\Large -4}}\;}{\;a^{{}^{\Large -2}}\,-\,b^{{}^{\Large -2}}\;}\phantom{X}$, com $\;{\small 0\neq a^2\neq b^2 \neq 0}$

resposta: $\,\dfrac{a^2+b^2}{a^2b^2}\,$
×

(FAAP) Simplificar $\phantom{X}\dfrac{\;ax^2\,-\,ay^2\;}{\;x^2\,-\,4xy\,+3y^2\;}\phantom{X}$ (obs.: supor x ≠ y)

resposta: $\,\frac{a(x+y)}{x-3y}\,$
×

$\phantom{X}x^{{}^{\Large 2m}}\,-\,1\phantom{X}$ é igual a:

$\,(x^m\,+\,1)(x^m\,-\,1)\,$

$\,(x^m\,+\,1)^2\,$

$\,(x^m\,+\,1)(x\,-\,1)\phantom{X}$

$\,(x^m\,+\,1)(x^m\,-\,1)\,$

$\,(x^m\,-\,1)^2\,$

resposta: (A)
×

(ENG ARARAQUARA) Simplificar $\phantom{X}\left( \dfrac{\;1\,-\,a\;}{a}\right)\,\div\,\left( 1\,-\,\dfrac{1}{\;a^2\;} \right)\phantom{X}$

resposta: $\,-\frac{a}{a+1}\,$
×

(FEI MAUÁ) Supondo $\phantom{X}x\phantom{X}$e$\phantom{X}y\phantom{X}$ reais com $\;x\,-\,y\,\neq\,0\;$ e $\;x\,+\,y\,\neq\,0\;$ simplificar a expressão algébrica $\phantom{X}\dfrac{\;x^3\,-\,y^3\;}{x\,-\,y}\,-\,\dfrac{\;x^3\,+\,y^3\;}{x\,+\,y}\phantom{X}$.

resposta: $\,2xy\,$
×

(F E EDSON DE QUEIROZ) Se $\phantom{X}M\,=\,a\,+\,\dfrac{\;b\,-\,a\;}{\;1\,+\,ab}\phantom{X}$e$\phantom{X}N\,=\,1\,-\,\dfrac{\;ab\,-\,a^2\;}{1\,+\,ab}\phantom{X}$, com $\;ab\,\neq\,1\;$, então $\;\dfrac{M}{N}\;$ é:

1 + ab

a - b

1 - ab

resposta: (B)
×

(USP) Uma expressão equivalente a $\phantom{X}2\,+\,\sqrt{\;\dfrac{\,a^2\,}{\,b^2\,}\,+\,\dfrac{\,b^2\,}{\,a^2\,}\,+\,2\phantom{X}}\phantom{X}$, para a > 0 e b > 0 é:

$\,\dfrac{\,a\,+\,b\,}{ab}\;\phantom{XX}$

$\,\dfrac{\,(a\,+\,b)^2\,}{ab}\,$

$\,\left(\dfrac{\,a\,+\,b\,}{ab}\right)^2\,$

$\,a^2\,+\,b^2\,+\,2ab\,$

$\,a\,+\,b\,+\,2\,$

resposta: (B)
×

(USP) Simplificando a expressão $\phantom{X}\left(\dfrac{\,a\,+\,b\,}{\,a\,-\,b\,}\,-\,\dfrac{\,a\,-\,b\,}{\,a\,+\,b\,} \right)\,\centerdot\,\dfrac{\,a\,+\,b\,}{\,2ab\,}\phantom{X}$ obtém-se:
(Observação: supor a ≠ b, a ≠ -b, ab ≠ 0.)

$\,\dfrac{1}{\,b\,-\,a\,}\,$

$\,\dfrac{2}{\,a\,-\,b\,}\,$

$\,\dfrac{\,a\,-\,b\,}{2}\,$

$\,\dfrac{1}{\,2ab\,}\,$

não sei

resposta: (B)
×

(UFGO) Simplificando a expressão $\phantom{X}\dfrac{\;a^2\,+\,a\;}{\;b^2\,+\,b\;}\centerdot \dfrac{\;a^2\,-\,a\;}{\;b^2\,-\,b\;}\centerdot \dfrac{\;b^2\,-\,1\;}{\;a^2\,-\,1\;}\phantom{X}$ obtém-se:

$\,\dfrac{\;a\;}{\;b\;}\phantom{X}$

$\,\dfrac{\;b\;}{\;a\;}\,$

$\,\dfrac{\;a^2\;}{\;b^2\;}\,$

$\,\dfrac{\;b^2\;}{\;a^2\;}\,$

$\,\dfrac{\;ab\;}{\;b\;}\,$

Obs.: Supor a ≠ 1, a ≠ -1, b ≠ 1, b ≠ -1, b ≠ 0

resposta: (C)
×

Determinar em $\,\mathbb{R}\,$ o conjunto verdade das equações $\phantom{X}3x\,-\,\left[\,2\,-\,(x\,-\,1)\,\right]\,=\,5x\phantom{X}$

resposta: $\,3x\,-\,\left[\,2\,-\,(x\,-\,1)\,\right]\,=\,5x\; \Leftrightarrow$ $\,3x\,-\,\left[\,2\,-\,x\,+\,1\right]\,=\,5x\;\Leftrightarrow$ $\,3x\,-\,2\,+\,x\,-\,1\,=\,5x\; \Leftrightarrow$ $\,3x\,+\,x\,-\,5x\,=\,2\,+\,1\;\Leftrightarrow$ $\,-x\,=\,3\;\Leftrightarrow\;x\,=\,-3\phantom{XX}$ V = {-3}
×

Determinar o conjunto solução do sistema $\,\left\{\begin{array}{rcr} 4x\,+\,10y\,=\,2\phantom{X} & \\ -3x\,-\,2y\,=\,4\phantom{X} & \\ \end{array} \right.\,$

resposta: S = {-2; 1}
×

Determinar em $\,\mathbb{R}\,$ o conjunto verdade das seguintes equações:

$\,2x^2\,-\,8\,=\,0\,$

$\,x^2\,-\,2x\,=\,0\,$

$\,5x^2\,=\,0\,$

resposta: a) V = {2; -2} b) V = {0; 2} c) V = {0}
×

(UFGO) Simplificando $\phantom{X}\dfrac{\;(x\,+\,y)^3\,-\,2y(y\,+\,x)^2\;}{x^2\,-\,y^2}\phantom{X}$ temos:

$\,\dfrac{\,(x\,+\,y)^2\,}{x\,-\,y}$

$\,x\,-\,y\,-\,2yx^2\,$

$\;x\,+\,y\phantom{XX}$

$\,x\,-\,y\,$

$\,\dfrac{\;x^2\,+\,y^2\;}{x\,-\,y}$

Observação: supor x ≠ y e x ≠ -y.

resposta: (C)
×

(PUC) Simplificando a expressão $\phantom{X}\dfrac{\;2(x\,-\,2)(x\,-\,3)^3\,-\,3(x\,-\,2)^2(x\,-\,3)^2\;}{(x\,-\,3)^6}\phantom{X}$ obtém-se:

$\,\dfrac{\;x(x\,-\,2)\;}{(x\,-\,3)^3}\,$

$\,\dfrac{\;x(2\,-\,x)\;}{(x\,-\,3)^3}\,$

$\,\dfrac{\;x(x\,-\,2)\;}{(x\,-\,3)^4}\,$

$\,\dfrac{\;x(2\,-\,x)\;}{(x\,-\,3)^4}\,$

$\,\dfrac{\;5x(x\,-\,2)\;}{(x\,-\,3)^4}\,$

Observação: supor x ≠ 3

resposta: (D)
×

(USP) A expressão $\phantom{X}\dfrac{\;x^2\,-\,1\;}{x^2}\,\div\,\dfrac{x^2\,-\,2x\,-\,3}{\;x^3\,-\,6x^2\,+\,9x\;}\phantom {X}$ é equivalente, para valores de x que não anulam nenhum dos 4 polinômios citados, a

$\,x\,-\,4\,+\,\frac{3}{x}\;\;$

$\,x\,-\,2\,+\,\frac{3}{x}\,$

$\,x^2\,-\,4x\,+\,3$

$\,x^2\,-\,3x\phantom{x \over x}$

$\,x^3\,-\,2x^2\,-\,3x\,$

resposta: (A)
×

(PUC) Simplificada a expressão $\phantom{X}\dfrac{\;x^3\,-\,3x^2y^2\,+\,2xy^3\;}{x^4y\,-\,8xy^4}\phantom{X}$ temos:

$\,\dfrac{x\,-\,y}{\;x^2\,+\,2xy\,+\,4y^2\;}\,$

$\,\dfrac{x\,+\,y}{\;x\,-\,y\;}\,$

$\,\dfrac{x(x\,-\,y)}{\;x(x\,+\,y)\;}\,$

$\,\dfrac{x\,-\,y}{\;(x\,-\,2y)^2\;}\,$

$\,\dfrac{x\,+\,2y}{\;x^2\,-\,2x\,+\,4y^2\;}\,$

resposta: (A)
×

(UBERABA) Racionalizando-se o denominador da fração $\phantom{X}\dfrac{2\sqrt{3}}{\;\sqrt{5}\,-\,\sqrt{3}\;}\phantom{X}$ obtém-se:

$\,\dfrac{\;\sqrt{15}\,-\,\sqrt{3}\;}{2}\,$

$\,\dfrac{\;\sqrt{15}\,+\,\sqrt{3}\;}{2}\,$

$2(\sqrt{15}\,+\,\sqrt{3})\,$

$\,\sqrt{15}\,+\,3\,$

$\,\sqrt{15}\,-\,3\,$

resposta: (D)
×

(MACKENZIE) $\phantom{X}\dfrac{1}{\;1\,-\,\sqrt{2}\;}\,-\,\dfrac{1}{\;\sqrt{2}\,+\,1\;}\phantom{X}$ é igual a:

$\,\sqrt{2}\;\;$

$\,-2\phantom{X}$

$\,2\;\phantom{X}\,$

$\,2(\sqrt{2}\,+\,1)\,$

$\,-2\sqrt{2}\,$

resposta: (E)
×

(FAAP) Simplificar $\phantom{X}\dfrac{\;2\,+\,\sqrt{3}\;}{\;1\,-\,\sqrt{5}\;}\,+\,\dfrac{\;2\,-\,\sqrt{3}\;}{\;1\,+\sqrt{5}\;}\phantom{X}$

resposta: $\,-1\,-\,\dfrac{\sqrt{15}}{2}\,$
×

(UNB) Sejam a e b reais com |a| > |b|. O valor da expressão $\phantom{X}\dfrac{\;\sqrt{a^2\,-\,b^2\,}}{a\,-\,b}\phantom{X}$ é:

$\,ab\,$

$\,\sqrt{\,a\,+\,b\;}$

$\,\sqrt{\;\dfrac{\;a\,+\,b\;}{a\,-\,b\;}\;}\,$

nenhuma dessas

resposta: (E)
×

(UNB) Se $\phantom{X}P\,=\,\dfrac{1}{\,\sqrt{\,7\;}\,-\,\sqrt{\,5\;}\;}\phantom{X}$; $\phantom{X}Q\,=\,\dfrac{1}{\,\sqrt{\,8\;}\,-\,\sqrt{\,5\;}\;}\phantom{X}$; $\phantom{X}R\,=\,\dfrac{\,\sqrt{\,5\;}\,+\,\sqrt{\,8\;}\;}{3}\phantom{X}$ então:

P < Q < R

P > Q, Q < R

P > Q > R

P > Q = R

P = Q < R

resposta: (D)
×

(FEI MAUÁ) Racionalizar o denominador da fração $\phantom{X}\dfrac{1}{\;1\,+\,\sqrt{\,2\;}\,-\,\sqrt{\,3\;}\;}\phantom{X}$

resposta: $\,\dfrac{\;\sqrt{\,2\;}(1\,+\,\sqrt{\,2\;}\,+\,\sqrt{\,3\;})\;}{4}\,$
×

(PUC) O número $\phantom{X}3\,+\,\sqrt{\,3\;}\,+\,\dfrac{1}{\,3\,-\,\sqrt{\,3\;}\,}\,-\,\dfrac{1}{\,3\,+\,\sqrt{\,3\;}\,}\phantom{X}$ é igual a:

$\,3\,+\,\dfrac{\;3\sqrt{\,3\;}}{2}\,$

$\,3\,+\,\dfrac{\;4\sqrt{\,3\;}}{3}\,$

$\,3\,-\,\dfrac{\;4\sqrt{\,3\;}}{3}\,$

$\,\dfrac{\;3\,+\,2\sqrt{\,3\;}}{3}\,$

$\,\dfrac{\;3\,-\,2\sqrt{\,3\;}}{3/6}\,$

resposta: (B)
×

(UFSM) Efetuando o produto $\phantom{X}\dfrac{\;2\,-\,\sqrt{\,3\;}\;}{\sqrt{\,5\;}}\,\centerdot \,\dfrac{\;\sqrt{\,3\;}\,+\,2\;}{\sqrt{\,5\;}\,-\,1}\phantom{X}$ teremos:

$\,\dfrac{1}{\;5\,+\,\sqrt{\,5\;}\;}\phantom{X}$

$\,\dfrac{\;5\,+\,\sqrt{\,5\;}\;}{20}\,$

$\,\dfrac{1}{\;25\,-\,\sqrt{\,5\;}\;}\,$

$\,\dfrac{\;4\,-\,2\sqrt{\,3\;}}{\;2\sqrt{\,5\;}\,-\,1\;}\,$

nenhuma das anteriores

resposta: (B)
×