a) \(ĐKXĐ:x\ne4\)
\(P=\frac{3}{\sqrt{x}+1}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{9}{x-\sqrt{x}-2}\)
\(=\frac{3}{\sqrt{x}+1}-\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3.\left(\sqrt{x}-2\right)-\sqrt{x}.\left(\sqrt{x}+1\right)+9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-x+2\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{3-\sqrt{x}}{\sqrt{x}-2}\)
Vậy \(P=\frac{3-\sqrt{x}}{\sqrt{x}-2}\left(x\ne4\right)\)
Để \(P=1\Leftrightarrow\frac{3-\sqrt{x}}{\sqrt{x}-2}=1\)
\(\Rightarrow3-\sqrt{x}=\sqrt{x}-2\)
\(\Leftrightarrow2\sqrt{x}=5\Leftrightarrow\sqrt{x}=\frac{5}{2}\Leftrightarrow x=\frac{25}{4}\) ( Thỏa mãn ĐKXĐ )
Vậy \(x=\frac{25}{4}\frac{5}{4}\) thì \(P=1\)
a) ĐKXĐ: \(x\ne4\)
Ta có: \(\frac{3}{\sqrt{x}+1}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{9}{x-\sqrt{x}-2}\)
\(=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\frac{9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3\sqrt{x}-6-x-\sqrt{x}+9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3-x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3-\sqrt{x}}{\sqrt{x}-2}\)
b) Đặt P=1
\(\Leftrightarrow3-\sqrt{x}=\sqrt{x}-2\)
\(\Leftrightarrow3-\sqrt{x}-\sqrt{x}+2=0\)
\(\Leftrightarrow5-2\sqrt{x}=0\)
\(\Leftrightarrow2\sqrt{x}=5\)
\(\Leftrightarrow\sqrt{x}=\frac{5}{2}\)
\(\Leftrightarrow x=\frac{25}{4}\)(tm)
