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Đáp án:
a) $\dfrac{\left( x-2 \right)\left( 2x+2{{x}^{2}} \right)}{\left( x+1 \right)\left( 4x-{{x}^{3}} \right)}=\dfrac{-2}{x+2}$
b) $\dfrac{{{x}^{2}}+{{y}^{2}}+2xy-1}{{{x}^{2}}-{{y}^{2}}+1+2x}=\dfrac{x+y-1}{x+1-y}$
Giải thích các bước giải:
a) $\dfrac{\left( x-2 \right)\left( 2x+2{{x}^{2}} \right)}{\left( x+1 \right)\left( 4x-{{x}^{3}} \right)}=\dfrac{-2}{x+2}$
$VT=\dfrac{\left( x-2 \right)\left( 2x+2{{x}^{2}} \right)}{\left( x+1 \right)\left( 4x-{{x}^{3}} \right)}$
$=\dfrac{\left( x-2 \right).2x\left( 1+x \right)}{\left( x+1 \right).x\left( 4-{{x}^{2}} \right)}$
$=\dfrac{2\left( x-2 \right)}{\left( 2-x \right)\left( 2+x \right)}$
$=\dfrac{-2}{x+2}=VP$
Vậy $\dfrac{\left( x-2 \right)\left( 2x+2{{x}^{2}} \right)}{\left( x+1 \right)\left( 4x-{{x}^{3}} \right)}=\dfrac{-2}{x+2}$
b) $\dfrac{{{x}^{2}}+{{y}^{2}}+2xy-1}{{{x}^{2}}-{{y}^{2}}+1+2x}=\dfrac{x+y-1}{x+1-y}$
$VT=\dfrac{{{x}^{2}}+{{y}^{2}}+2xy-1}{{{x}^{2}}-{{y}^{2}}+1+2x}$
$=\dfrac{\left( {{x}^{2}}+2xy+{{y}^{2}} \right)-1}{\left( {{x}^{2}}+2x+1 \right)-{{y}^{2}}}$
$=\dfrac{{{\left( x+y \right)}^{2}}-{{1}^{2}}}{{{\left( x+1 \right)}^{2}}-{{y}^{2}}}$
$=\dfrac{\left( x+y+1 \right)\left( x+y-1 \right)}{\left( x+1+y \right)\left( x+1-y \right)}$
$=\dfrac{x+y-1}{x+1-y}=VP$
Vậy $\dfrac{{{x}^{2}}+{{y}^{2}}+2xy-1}{{{x}^{2}}-{{y}^{2}}+1+2x}=\dfrac{x+y-1}{x+1-y}$
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