Đáp án: $A=\dfrac{-2\left(\sqrt{x}-1\right)}{\sqrt{x}}$
Giải thích các bước giải:
Ta có:
$A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot \dfrac{\left(1-x\right)^2}{x}$
$\to A=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot \dfrac{\left(x-1\right)^2}{x}$
$\to A=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot \dfrac{\left(\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right)^2}{x}$
$\to A=\left(\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot \dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{x}$
$\to A=\left(\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot \dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{x}$
$\to A=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot \dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{x}$
$\to A=\dfrac{-2\left(\sqrt{x}-1\right)}{\sqrt{x}}$
