Lista de exercícios do ensino médio para impressão
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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
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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)
×
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 $\phantom{X}\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\phantom{X}$

 



resposta:

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


Resolução:
Multiplicamos o numerador e o denominador da fração por $\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,$
$\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\;=$ $\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\,\centerdot\,\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;\left[\,2\,+\,\sqrt{3}\,+\,\sqrt{7}\,\right]\,\centerdot\,\left[\,\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,\right] \;}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;(2\,+\sqrt{3\,})^2\,-\,(\sqrt{7\,})^2\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\,+\,2\,\centerdot\,2\,\centerdot\,\sqrt{3\,}\,+\,3\,-\,7\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,\centerdot\,\dfrac{\;\sqrt{3\,}\;}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\sqrt{3\,}\,+\,(\sqrt{3\,})^2\,-\,\sqrt{3\,} \centerdot \sqrt{7\,}\,}{(\sqrt{3\,})^2}\,=$ $\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;$
$\boxed{\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;}$
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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\,} $

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

×
(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)
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(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)
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(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)
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Veja exercÍcio sobre:
álgebra elementar
fatoração
diferença de quadrados