a.
\(P=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\dfrac{x-1}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{1}{x+\sqrt{x}+1}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{1}{x-1}\)
b.
Đặt \(Q=x^2.P=x^2.\dfrac{1}{x-1}=\dfrac{x^2}{x-1}=\dfrac{x^2-1+1}{x-1}=\dfrac{\left(x-1\right)\left(x+1\right)+1}{x-1}=x+1+\dfrac{1}{x-1}\)
Để Q nguyên \(\Rightarrow\dfrac{1}{x-1}\in Z\)
\(\Rightarrow x-1\inƯ\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{0;2\right\}\)
