a: ĐKXĐ: x>=0 và x<>1
\(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\cdot\sqrt{x}}\right)\cdot\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Khi \(x=8-2\sqrt{7}=\left(\sqrt{7}-1\right)^2\) thì
\(P=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}+1}{\sqrt{\left(\sqrt{7}-1\right)^2}-1}\)
\(=\dfrac{\sqrt{7}-1+1}{\sqrt{7}-1-1}=\dfrac{\sqrt{7}}{\sqrt{7}-2}=\dfrac{\sqrt{7}\left(\sqrt{7}+2\right)}{3}\)
e: P=6/5
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{6}{5}\)
=>\(6\left(\sqrt{x}-1\right)=5\left(\sqrt{x}+1\right)\)
=>\(6\sqrt{x}-6=5\sqrt{x}+5\)
=>\(6\sqrt{x}-5\sqrt{x}=6+5\)
=>\(\sqrt{x}=11\)
=>x=121(nhận)
