a) Ta có: \(P=\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}-\frac{2\sqrt{x}}{4-x}\)
\(=\frac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{2+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\frac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\frac{-3\sqrt{x}+6}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\frac{3}{2+\sqrt{x}}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Để \(P=\frac{6}{5}\) thì \(\frac{3}{2+\sqrt{x}}=\frac{6}{5}\)
\(\Leftrightarrow6\cdot\left(2+\sqrt{x}\right)=15\)
\(\Leftrightarrow12+6\sqrt{x}=15\)
\(\Leftrightarrow6\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)
hay \(x=\frac{1}{4}\)(nhận)
Vậy: để \(P=\frac{6}{5}\) thì \(x=\frac{1}{4}\)
c) Để P nhận giá trị nguyên thì \(3⋮\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)
\(\Leftrightarrow\sqrt{x}+2\in\left\{1;-1;3;-3\right\}\)
mà \(\sqrt{x}+2\ge2\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+2=3\)
\(\Leftrightarrow\sqrt{x}=1\)
hay x=1(nhận)
Vậy: Để P nhận giá trị nguyên thì x=1
