Bài 33:
a: ĐKXĐ: x>0; x<>1
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
b: P<0
=>x-1<0
=>x<1
=>0<x<1
c: Thay \(x=4-2\sqrt3=\left(\sqrt3-1\right)^2\) vào P, ta được:
\(P=\frac{4-2\sqrt3-1}{\sqrt{\left(\sqrt3-1\right)^2}}=\frac{3-2\sqrt3}{\sqrt3-1}\)
\(=\frac{\left(3-2\sqrt3\right)\left(\sqrt3+1\right)}{2}=\frac{3\sqrt3+3-6-2\sqrt3}{2}=\frac{\sqrt3-3}{2}\)
Bài 34:
a: ĐKXĐ: x>=0; x<>1
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}:\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-4}=\frac{\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)
b: \(P=\frac12\)
=>\(\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac12\)
=>\(2\sqrt{x}-2=\sqrt{x}+2\)
=>\(\sqrt{x}=4\)
=>x=16(Nhận)
