Question

P=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)

Answer 1

ĐKXD:\(x\ne1;x\ge0\)

\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(1-x\right)^2}{2}\)

\(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(\Leftrightarrow P=\dfrac{x-2\sqrt{x}+\sqrt{x}-2-x-2\sqrt{x}+\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{x-1}{2}\)

\(\Rightarrow P=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{x-1}{2}\)

=>P=\(P=\sqrt{x}\left(1-\sqrt{x}\right)\)