Trả lời:
a, \(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)
\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(x+\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(x+2\sqrt{x}-\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left[\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)\right]\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\) (đpcm)
b, \(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}}=2\sqrt{x}+5\) \(\left(ĐK:x>0\right)\)
\(\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)
\(\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=\frac{2x}{\sqrt{x}}+\frac{5\sqrt{x}}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)
\(\Leftrightarrow2x+3\sqrt{x}-2=0\)
\(\Leftrightarrow2x+4\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=0\\2\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-2\left(voli\right)\\2\sqrt{x}=1\end{cases}\Leftrightarrow}\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)}\)
Vậy x = 1/4 là giá trị cần tìm.
