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`Answer:`
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\left(ĐK:x>0;x\ne9;x\ne25\right)\)
\(=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=-\frac{3\sqrt{x}-x+2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)
\(=-\frac{\sqrt{x}\left(3+\sqrt{x}\right)}{3+\sqrt{x}}.\frac{\sqrt{x}}{5-\sqrt{x}}\)
\(=-\sqrt{x}.\frac{\sqrt{x}}{5-\sqrt{x}}\)
\(=\frac{x}{\sqrt{x}-5}\)
p=\(\frac{-\sqrt{x}}{2\left(x-\sqrt{x}+1\right)}\)
mà \(-\sqrt{x}< 0\) ( vì điều kiện xác định x > 0 ; x \(\ne\) -1 )
mặt khác \(x-\sqrt{x}+1=x-2\sqrt{x}\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=> \(\frac{-\sqrt{x}}{2\left(x-\sqrt{x}+1\right)}< 0\)
=> p \(< \) 0 => p luôn luôn âm với mọi x
p = \(\frac{1}{\sqrt{x}+1}-\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p = \(\frac{x-\sqrt{x}+1}{\left(x-\sqrt{x}+1\right)\sqrt{x}+1}-\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p = \(\frac{x-\sqrt{x}+1-x-2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p=\(\frac{-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p=\(\frac{-1}{\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p=\(\frac{-\sqrt{x}}{2\left(x-\sqrt{x}+1\right)}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}-\frac{2\sqrt{x}-1}{\sqrt{x}+2}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)
\(A=\frac{\sqrt{x}-2}{x-4}\)