Giúp em với ạ em xik cảm ơn nhìu

Đáp án:

a) $A=\dfrac{4}{x-1}$

b) $x=3$

Giải thích các bước giải:

a)

$A=\dfrac{5x+1}{{{x}^{3}}-1}-\dfrac{2}{1-x}+\dfrac{2x-1}{{{x}^{2}}+x+1}\,\,\,\left( x\ne 1 \right)$

$A=\dfrac{5x+1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}+\dfrac{2}{x-1}+\dfrac{2x-1}{{{x}^{2}}+x+1}$

$A=\dfrac{5x+1+2\left( {{x}^{2}}+x+1 \right)+\left( x-1 \right)\left( 2x-1 \right)}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}$

$A=\dfrac{5x+1+2{{x}^{2}}+2x+2+2{{x}^{2}}-x-2x+1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}$

$A=\dfrac{4{{x}^{2}}+4x+4}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}$

$A=\dfrac{4\left( {{x}^{2}}+x+1 \right)}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}$

$A=\dfrac{4}{x-1}$

b)

Để $A=2$

Thì $\dfrac{4}{x-1}=2$

$\Rightarrow 2\left( x-1 \right)=4$

$\Rightarrow x-1=2$

$\Rightarrow x=3$ (thỏa mãn)

Vậy $x=3$ thì $A=2$