Lista de exercícios do ensino médio para impressão
Fatore:
a)
$\,12a^3b^2\,-\,30a^2b^3\,$
b)
$\,xy\,+\,3x\,+\,4y\,+\,12\,$
c)
$\,6ab\,+\,4b^3\,+\,15a^3\,+\,10a^2b^2\,$
d)
$\,a^2 \,-\, 25\,$
e)
$\,x^2\,-\,1\,$
f)
$\,x^4\,-\,1\,$
g)
$\,144\,-\,81a^2b^2\,$

 



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) $\,9(4+3ab) \centerdot (4-3ab)\,$

×
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:
a.
$\phantom{X}(2\,+\,3m)^{\large 2}\phantom{X}$
b.
$\phantom{X}(a\,-\,3)^{\large 2}\phantom{X}$
c.
$\phantom{X}(\sqrt{5}\,+\,\sqrt{3})^{\large 2}\phantom{X}$
d.
$\phantom{X}(a\,+\,3b)^{\large 3}\phantom{X}$
e.
$\phantom{X}(2a\,-\,b)^{\large 3}\phantom{X}$

 



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:
a.
$\phantom{X}a^{\large 2}\,+\,4a\,+\,4\phantom{X}$
b.
$\phantom{X}9a^{\large 2}\,+\,30ab\,+\,25b^{\large 2}\phantom{X}$
c.
$\phantom{X}1\,-\,18x^{\large 2}\,+\,81x^{\large 4}\phantom{X}$
d.
$\phantom{X}a^{\large 3}\,+\,27\phantom{X}$
e.
$\phantom{X}64\,-\,x^{\large 3}\phantom{X}$

 



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.
a)
$\;\dfrac{\;x^2\,+\,2xy\,+\,y^2}{x^2\,-\,y^2}\;$
b)
$\;\dfrac{a^3\,-\,1}{a^2\,-\,1}\;$
c)
$\;\dfrac{m^3\,+\,n^3}{m^3\,-\,m^2n\,+\,mn^2}\;$
d)
$\;\dfrac{x^3\,+\,3x^2y\,+\,3xy^2\,+\,y^3}{x^3\,+\,y^3}\,\div\,\dfrac{x^2\,+\,2xy\,+\,y^2}{x^2\,-\,xy\,+\,y^2}\;$

 



resposta:
a)
$\,\frac{x+y}{x-y}\,$
b)
$\,\frac{a^2+a+1}{a+1}\,$
c)
$\,\frac{m+n}{m}$
d)
1

×
Fatorar $\phantom{X}a^3\,-\,8\phantom{X}$

 



resposta: $\,a^3\,-\,8\,=\,a^3\,-\,2^3\,=$ $\,(a\,-\,2)(a^2\,+\,a\centerdot 2\,+\,2^2)\,=$ $\,(a\,-\,2)(a^2\,+\,2a\,+4)\,$
×
Fatorar: $\phantom{X}x^3\,+\,1\phantom{X}$

 



resposta: $\,x^3\,+\,1\,=\,x^3\,+\,1^3\,=$ $\,(x\,+\,1)(x^2\,-\,a\centerdot 1\,+\,1^2)\,=$ $\,(x\,+\,1)(x^2\,-\,x\,+1)\,$
×
Fatorar: $\phantom{X}x^3\,+\,2x^2\,+\,2x\,+1\phantom{X}$

 



resposta: Resolução:
$\,x^3\,+\,2x^2\,+\,2x\,+1\,=$ $\,x^3\,+\,1\,+\,2x^2\,+\,2x\,=$ $\,(x^3\,+\,1)\,+\,2x(x\,+\,1)\,=$ $\,(x\,+\,1)(x^2\,-\,x\,+\,1)\,+\,2x(x\,+\,1)\,=$ $\,(x\,+\,1)\left[\,(x^2\,-\,x\,+\,1)\,+\,2x\,\right]\,=$
$\,(x\,+\,1)(x^2\,+\,x\,+\,1)$
×
Sabendo-se que $\phantom{X}a\,+\,\dfrac{1}{a}\,=\,3\phantom{X}$, calcular o valor de $\phantom{X}a^3\,+\,\dfrac{1}{a^3}\phantom{X}$

 



resposta: Resolução:
$\,a\,+\,\dfrac{1}{a}\,=\,3\;\Rightarrow$ $\,(a\,+\,\dfrac{1}{a})^3\,=\,3^3\;\Leftrightarrow$ $\,a^3\,+\,3a\,+\,\dfrac{3}{a}\,+\,\dfrac{1}{a^3}\,=\,27\;\Leftrightarrow$ $\,a^3\,+\,\dfrac{1}{a^3}\,+\,3a\,+\,\dfrac{3}{a}\,=\,27\;\Leftrightarrow$ $\,a^3\,+\,\dfrac{1}{a^3}\,+\,3(a\,+\,\dfrac{1}{a})\,=\,27\;\Leftrightarrow$ $\,a^3\,+\,\dfrac{1}{a^3}\,+\,3\,\centerdot \,3\,=\,27\;\Leftrightarrow$
$\,a^3\,+\,\dfrac{1}{a^3}\,=\,27\,-\,9\,=\,18\,$

×
Racionalizar o denominador da fração $\phantom{X}\dfrac{\sqrt{2\,}}{\;2\,+\,\sqrt{3\,}\;}\phantom{X}$

 



resposta: Resolução:
Sabendo que (a + b)(a - b) = a² - b², para racionalizar o denominador da fração acima devemos multiplicar o numerador e o denominador pelo valor $\;2\,-\,\sqrt{3}\;$
$\dfrac{\sqrt{2}}{\;2\,+\,\sqrt{3}\;}\;=\;\dfrac{\sqrt{2}}{\;2\,+\,\sqrt{3}\;}\,\centerdot\,\dfrac{2\,-\,\sqrt{3}}{\;2\,-\,\sqrt{3}\;}\,=\,\dfrac{\;\sqrt{2}(2\,-\,\sqrt{3})\;}{2^2\,-\,(\sqrt{3})^2}\,=$ $\,\dfrac{\;\sqrt{2}(2\,-\,\sqrt{3})\;}{4\,-\,3}\;=\;\dfrac{\;\sqrt{2}(2\,-\,\sqrt{3})\;}{1}\;=\;\sqrt{\,2\,}\,(2\,-\,\sqrt{\,3\,})\,=$
$\,2\sqrt{2\,}\,-\,\sqrt{6\,}$

×
Racionalizar o denominador da fração $\phantom{X}\dfrac{2}{\;\sqrt{5\,}\,-\,\sqrt{3\,}\;}\phantom{X}$

 



resposta: Resolução:
Sabendo que (a + b)(a - b) = a² - b², para racionalizar o denominador da fração acima devemos multiplicar o numerador e o denominador pelo valor $\;\sqrt{5\,}\,+\,\sqrt{3}\;$
$\dfrac{2}{\;\sqrt{5\,}\,-\,\sqrt{3}\;}\;=\;\dfrac{2}{\;\sqrt{5\,}\,-\,\sqrt{3}\;}\,\centerdot\,\dfrac{\sqrt{5\,}\,+\,\sqrt{3}}{\;\sqrt{5\,}\,+\,\sqrt{3}\;}\,=$ $\,\dfrac{\;2(\sqrt{5\,}\,+\,\sqrt{3})\;}{(\sqrt{5\,})^2\,-\,(\sqrt{3})^2}\,=$ $\,\dfrac{\;2(\sqrt{5\,}\,+\,\sqrt{3})\;}{5\,-\,3}\;=\;\dfrac{\;2(\sqrt{5\,}\,+\,\sqrt{3})\;}{2}\;=$
$\;\sqrt{\,5\,} + \sqrt{3\,} $

×
Racionalizar o denominador da fração $\phantom{X}\dfrac{1}{\sqrt[\Large 3]{5}\,+\,\sqrt[\Large 3]{2}}\phantom{X}$

 



resposta:

SOMA E DIFERENÇA DE CUBOS

$\,\boxed{\;a^3\,+\,b^3\,=\,(a\,+\,b)\,\centerdot\,(a^2\,-\,ab\,+\,b^2)\,}$
$\,\boxed{\;a^3\,-\,b^3\,=\,(a\,-\,b)\,\centerdot\,(a^2\,+\,ab\,+\,b^2)\,}$

Resolução:
Devemos multiplicar o numerador e o denominador da fração pelo fator que complete a expressão do produto notável correspondente.
$\,\dfrac{1}{\sqrt[\Large 3]{5}\,+\,\sqrt[\Large 3]{2}}\,=$ $\,\dfrac{1}{\sqrt[\Large 3]{5}\,+\,\sqrt[\Large 3]{2}}\,\centerdot\,\dfrac{(\sqrt[\Large 3]{5})^2\,-\,\sqrt[\Large 3]{5}\centerdot\sqrt[\Large 3]{2}\,+\,(\sqrt[\Large 3]{2})^2}{(\sqrt[\Large 3]{5})^2\,-\,\sqrt[\Large 3]{5}\centerdot\sqrt[\Large 3]{2}\,+\,(\sqrt[\Large 3]{2})^2}\,=$ $\,\dfrac{\sqrt[\Large 3]{5^2}\,-\,\sqrt[\Large 3]{5\,\centerdot \,2\;}\,+\,\sqrt[\Large 3]{2^2}}{(\sqrt[\Large 3]{5\;})^3\,+\,(\sqrt[\Large 3]{2\;})^3}\,=$
$\,\dfrac{\;\sqrt[\Large 3]{25\;}\,-\;\sqrt[\Large 3]{10\;}\,+\,\sqrt[\Large 3]{4\;}\,}{7}\;$

×
Racionalizar o denominador da fração $\phantom{X}\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\phantom{X}$

 



resposta:

DIFERENÇA DE QUADRADOS
$\,\boxed{\;a^2\,-\,b^2\,=\,(a\,+\,b)\,\centerdot\,(a\,-\,b)\,}$


Resolução:
Multiplicamos o numerador e o denominador da fração por $\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,$
$\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\;=$ $\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\,\centerdot\,\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;\left[\,2\,+\,\sqrt{3}\,+\,\sqrt{7}\,\right]\,\centerdot\,\left[\,\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,\right] \;}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;(2\,+\sqrt{3\,})^2\,-\,(\sqrt{7\,})^2\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\,+\,2\,\centerdot\,2\,\centerdot\,\sqrt{3\,}\,+\,3\,-\,7\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,\centerdot\,\dfrac{\;\sqrt{3\,}\;}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\sqrt{3\,}\,+\,(\sqrt{3\,})^2\,-\,\sqrt{3\,} \centerdot \sqrt{7\,}\,}{(\sqrt{3\,})^2}\,=$ $\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;$
$\boxed{\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;}$
×
Sendo $\;a\;$ e $\;b\;$ números reais estritamente positivos e distintos, mostrar que $\phantom{X}\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,=\,\sqrt{a\,}\,+\,\sqrt{b\,}\phantom{X}$

 



resposta:

DIFERENÇA DE QUADRADOS
$\,\boxed{\;a^2\,-\,b^2\,=\,(a\,+\,b)\,\centerdot\,(a\,-\,b)\,}$


Resolução:
$\,\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,=$ $\,\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,\centerdot \,\dfrac{\sqrt{a\,}\,+\,\sqrt{b\,}}{\;\sqrt{a\,}\,+\,\sqrt{b\,}\;}\,=$ $\,\dfrac{\;(a\,-\,b)(\sqrt{a\,}\,+\,\sqrt{b\,})\;}{(\sqrt{a\,})^2\,-\,(\sqrt{b\,})^2}\,=$ $\,\dfrac{\;(a\,-\,b)(\sqrt{a\,}\,+\,\sqrt{b\,})\;}{(a\,-\,b)}\,=\,$$\,\sqrt{a\,}\,+\,\sqrt{b\,} $

×
Os denominadores das frações abaixo são diferentes de zero. Simplifique:
a) $\,\dfrac{\;a\,+\,a^2\;}{\;2\,+\,2a\;}$

b) $\,\dfrac{\;a^3\,+\,a^2b\;}{\;a^2\,+\,2ab\,+\,b^2\;}$

 



resposta: a) a/2b) $\,\frac{a^2}{a\,+\,b}$

×
Responda de acordo com o código
a)
Se todas forem verdadeiras
b)
Se I e II forem verdadeiras
c)
Se I e V forem verdadeiras
d)
Se II e IV forem verdadeiras
e)
Se todas forem falsas
I)
$\,2ap^2\,-\,32a\,=\,2a(p\,+\,4)(p\,-\,4)\,$
II)
$\,x^2\,-\,y^2\,-\,2y\,-\,1\,=\,(x\,+\,y\,+\,1)(x\,-\,y\,-\,1)\,$
III)
$\,\dfrac{x^2y^2}{4}\,-\,1\,=\,\left( \dfrac{x\,+\,y}{4}\,+\,1\right)\left( \dfrac{yx}{4}\,-\,1\right)\,$
IV)
$\,2(x\,-\,y)^3\,+\,x^3\,-\,y^3\,=\,3(x\,+\,y)(x^2\,-\,xy\,+\,y^2)\,$
V)
$\,x^{2m}\,-\,y^4\,=\,(x^m\,-\,y^2)(y^m\,-\,y)^2\,$

 



resposta: (B)
×
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 $\phantom{X}934287^2 - 934286^2\;$
a) 1868573b) 1975441c) 2d) 1e) nenhuma das anteriores
1868573197544121nenhuma das anteriores

 



resposta: (A)
×
A expressão $\phantom{X}\dfrac{\;2x\,-\,2y\;}{x\,-\,y}\phantom{X}$ é igual a 2 somente se:
a)
x > 0 e y < 0
b)
x ≠ 0 e y ≠ 0
c)
x ≠ y
d)
x, y ∈ $\mathbb{R}^{\large *}$
e)
todas são falsas

 



resposta: (C)
×
Sejam $\phantom{X}a\phantom{X}$e$\phantom{X}b\phantom{X}$ dois números reais diferentes de zero. A expressão $\phantom{X}\dfrac{1}{a^2} \,+\,\dfrac{2}{ab}\phantom{X}$é igual a:
a)
$\,\dfrac{b\,+\,2a}{a(a\,+\,b)}\,$
b)
$\,\dfrac{3}{a^2b}\,$
c)
$\,\dfrac{b\,+\,2a}{a^2b}\,$
d)
$\,\dfrac{b\,-\,2}{a(a\,+\,b)}\,$
e)
nenhuma dessas

 



resposta: (C)
×
(UNB) A expressão $\phantom{X}\dfrac{\;3a\,-\,4\;}{a^2\,-\,16}\,-\,\dfrac{1}{\;a\,-\,4\;}\phantom{X}$($\;{\small a\,\neq\,4}\;$) é equivalente a:
a)
$\,\dfrac{1}{\;a\,-\,4\;}\,$
b)
$\,\dfrac{2}{\;a\,+\,4\;}\,$
c)
$\,\dfrac{2}{\;a\,-\,4\;}\,$
d)
$\,\dfrac{\;a\,+\,4\;}{\;a\,-\,4\;}\,$
e)
nenhuma dessas

 



resposta: (B)
×
$\phantom{X}x^{{}^{\Large 2m}}\,-\,1\phantom{X}$ é igual a:
a)
$\,(x^m\,+\,1)(x^m\,-\,1)\,$
b)
$\,(x^m\,+\,1)^2\,$
c)
$\,(x^m\,+\,1)(x\,-\,1)\phantom{X}$
d)
$\,(x^m\,+\,1)(x^m\,-\,1)\,$
e)
$\,(x^m\,-\,1)^2\,$
 
 

 



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:
a) ab) bc) 1 + abd) a - be) 1 - ab
ab1 + aba - b1 - 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:
a)
$\,\dfrac{\,a\,+\,b\,}{ab}\;\phantom{XX}$
b)
$\,\dfrac{\,(a\,+\,b)^2\,}{ab}\,$
c)
$\,\left(\dfrac{\,a\,+\,b\,}{ab}\right)^2\,$
d)
$\,a^2\,+\,b^2\,+\,2ab\,$
e)
$\,a\,+\,b\,+\,2\,$
 
 

 



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.)
a)
$\,\dfrac{1}{\,b\,-\,a\,}\,$
b)
$\,\dfrac{2}{\,a\,-\,b\,}\,$
c)
$\,\dfrac{\,a\,-\,b\,}{2}\,$
d)
$\,\dfrac{1}{\,2ab\,}\,$
e)
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:
a)
$\,\dfrac{\;a\;}{\;b\;}\phantom{X}$
b)
$\,\dfrac{\;b\;}{\;a\;}\,$
c)
$\,\dfrac{\;a^2\;}{\;b^2\;}\,$
d)
$\,\dfrac{\;b^2\;}{\;a^2\;}\,$
e)
$\,\dfrac{\;ab\;}{\;b\;}\,$
 
 
Obs.: Supor a ≠ 1, a ≠ -1, b ≠ 1, b ≠ -1, b ≠ 0

 



resposta: (C)
×
(UFGO) Simplificando $\phantom{X}\dfrac{\;(x\,+\,y)^3\,-\,2y(y\,+\,x)^2\;}{x^2\,-\,y^2}\phantom{X}$ temos:
a)
$\,\dfrac{\,(x\,+\,y)^2\,}{x\,-\,y}$
b)
$\,x\,-\,y\,-\,2yx^2\,$
c)
$\;x\,+\,y\phantom{XX}$
d)
$\,x\,-\,y\,$
e)
$\,\dfrac{\;x^2\,+\,y^2\;}{x\,-\,y}$
 
 
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:
a)
$\,\dfrac{\;x(x\,-\,2)\;}{(x\,-\,3)^3}\,$
b)
$\,\dfrac{\;x(2\,-\,x)\;}{(x\,-\,3)^3}\,$
c)
$\,\dfrac{\;x(x\,-\,2)\;}{(x\,-\,3)^4}\,$
d)
$\,\dfrac{\;x(2\,-\,x)\;}{(x\,-\,3)^4}\,$
e)
$\,\dfrac{\;5x(x\,-\,2)\;}{(x\,-\,3)^4}\,$
Observação: supor x ≠ 3

 



resposta: (D)
×
(PUC) Simplificada a expressão $\phantom{X}\dfrac{\;x^3\,-\,3x^2y^2\,+\,2xy^3\;}{x^4y\,-\,8xy^4}\phantom{X}$ temos:
a)
$\,\dfrac{x\,-\,y}{\;x^2\,+\,2xy\,+\,4y^2\;}\,$
b)
$\,\dfrac{x\,+\,y}{\;x\,-\,y\;}\,$
c)
$\,\dfrac{x(x\,-\,y)}{\;x(x\,+\,y)\;}\,$
d)
$\,\dfrac{x\,-\,y}{\;(x\,-\,2y)^2\;}\,$
e)
$\,\dfrac{x\,+\,2y}{\;x^2\,-\,2x\,+\,4y^2\;}\,$

 



resposta: (A)
×
(ITA) Fatore a expressão:$\phantom{X}x^3\,+\,y^3\,+\,z^3\,-\,3xyz\phantom{X}$

 



resposta: $\,(x+y+z)(x^2+y^2+z^2-xy-xz-yz)\,$
×
(PUC - 1982) Sendo $\phantom{X}x^3\,+\,1\,=\,(x\,+\,1)(x^2\,+\,ax\,+\,b)\phantom{X}$ para todo $\phantom{X}x\phantom{X}$ real, os valores de a e b são respectivamente:
a)
-1 e -1
b)
0 e 0
c)
1 e 1
d)
1 e -1
e)
-1 e 1
 
 

 



resposta: (E)
×
(MED JUNDIAÍ - 1982) O valor numérico da expressão $\phantom{X}a^3\,-\,b^3\,+\,3ab^2\,-\,3a^2b\phantom{X}$ para $\phantom{X}a\,=\,\dfrac{\;\sqrt[\Large 3]{3\;}\,+\,2\;}{\sqrt[\Large 3]{\;2\;}}\phantom{X}$ e $\phantom{X}b\,=\,\dfrac{\;\sqrt[\Large 3]{3\;}\,-\,2\;}{\sqrt[\Large 3]{\;2\;}}\phantom{X}$ é:
a)
$\,\sqrt[\Large 3]{9\;}\,-\,2\;$
b)
$\,\sqrt[\Large 3]{9\;}\,+\,2\;$
c)
8
d)
13,5
e)
32

 



resposta: (E)
×
(MED JUNDIAÍ - 1982) Se $\,x\,$ e $\,y\,$ são números tais que $\phantom{X}y\,=\,\dfrac{x^2\,-\,3x\,+\,2}{\;x^3\,-\,2x^2\,-\,x\,+\,2\;}\phantom{X}$ então $\,y\,$ é igual a:
a)
$\,\dfrac{1}{\;x\,-\,1\;}\,$
b)
$\,\dfrac{1}{\;x\,+\,1\;}\,$
c)
$\,\dfrac{x\,-\,2}{\;x^2\,-\,1\;}\,$
d)
$\,\dfrac{x\,-\,1}{\;x\,-\,1\;}\,$
e)
$\,\dfrac{x\,+\,2}{\;(x\,+\,2)(x\,-\,1)\;}\,$

 



resposta: (B)
×
(FEI - 1982) Sejam os números $\phantom{X}a\,=\,\dfrac{\;\sqrt{5\;}\,-\,\sqrt{2\;}}{\;\sqrt{7\;}\,+\,\sqrt{3\;}}\phantom{X}$ e $\phantom{X}b\,=\,\dfrac{\;\sqrt{7\;}\,-\,\sqrt{3\;}}{\;\sqrt{5\;}\,+\,\sqrt{2\;}}\phantom{X}$.
Qual é o maior?

 



resposta: b > a
×
Veja exercÍcio sobre:
fatoração
álgebra elementar