Bài 6:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)
\(=\dfrac{x+2}{\sqrt{x}}\)
b: Thay \(x=3-2\sqrt{2}\) vào A, ta được:
\(A=\dfrac{3-2\sqrt{2}+2}{\sqrt{2}-1}=\dfrac{5-2\sqrt{2}}{\sqrt{2}-1}=3\sqrt{2}+1\)
Bài 5:
a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b: Để P<0 thì x-1<0
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1