1: Thay x=25 vào P, ta được:
\(P=\frac{\sqrt{25}+2}{\sqrt{25}-3}=\frac{5+2}{5-3}=\frac72\)
2:
a: \(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\)
\(=\frac{-\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-\left(\sqrt{x}+2\right)^2+\left(\sqrt{x}-2\right)^2-4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-x-4\sqrt{x}-4+x-4\sqrt{x}+4-4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-4x-8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-4\sqrt{x}}{\sqrt{x}-2}\)
\(Q=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\left(\frac{5}{2-\sqrt{x}}-\frac{4\sqrt{x}+3}{2\sqrt{x}-x}\right)\)
\(=\frac{-4\sqrt{x}}{\sqrt{x}-2}:\frac{5\sqrt{x}-4\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(=\frac{-4\sqrt{x}}{\sqrt{x}-2}\cdot\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\frac{4x}{\sqrt{x}-3}\)
b: \(\frac{P}{Q}=\frac52\)
=>\(\frac{\sqrt{x}+2}{\sqrt{x}-3}:\frac{4x}{\sqrt{x}-3}=\frac52\)
=>\(\frac{\sqrt{x}+2}{4x}=\frac52\)
=>\(20x=2\left(\sqrt{x}+2\right)=2\sqrt{x}+4\)
=>\(10x=\sqrt{x}+2\)
=>\(10x-\sqrt{x}-2=0\)
=>\(10x-5\sqrt{x}+4\sqrt{x}-2=0\)
=>\(\left(2\sqrt{x}-1\right)\left(5\sqrt{x}+2\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>\(2\sqrt{x}=1\)
=>\(\sqrt{x}=\frac12\)
=>x=1/4(nhận)
