\(1,x=25\Rightarrow P=\dfrac{\sqrt{25}-3}{\sqrt{25}+1}=\dfrac{5-3}{5+1}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(2,Q=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+2}{x-1}\left(dkxd:x>0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}+1\right)+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(3,A=\dfrac{P}{Q}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}:\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}.\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(A=2m\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}}=2m\Leftrightarrow\sqrt{x}-3-2m\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(1-2m\right)-3=0\)
Để pt A có nghiệm thì \(1-2m\ge0\Leftrightarrow-2m\ge-1\Leftrightarrow m\le\dfrac{1}{2}\)
