a) Ta có: \(P=\frac{\sqrt{x}}{\sqrt{x}+5}-\frac{\sqrt{x}+1}{5-\sqrt{x}}-\frac{5-9\sqrt{x}}{x-25}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5-9\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{x-5\sqrt{x}+x+6\sqrt{x}+5-5+9\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{2x+10\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{2\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}-5}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
Để P=3 thì \(\frac{2\sqrt{x}}{\sqrt{x}-5}=3\)
\(\Leftrightarrow2\sqrt{x}=3\left(\sqrt{x}-5\right)\)
\(\Leftrightarrow2\sqrt{x}=3\sqrt{x}-15\)
\(\Leftrightarrow2\sqrt{x}-3\sqrt{x}+15=0\)
\(\Leftrightarrow-\sqrt{x}+15=0\)
\(\Leftrightarrow-\sqrt{x}=-15\)
\(\Leftrightarrow\sqrt{x}=15\)
hay x=225(thỏa mãn)
Vậy: Khi P=3 thì x=225
