1: \(P=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
2: Để P=3 thì \(3\sqrt{x}-9=\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}=9\)
hay x=81/4
1, đk : x >= 0 ; x khác 9
\(P=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{x-9}\)
\(=\dfrac{x+3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
2, Ta có : \(\dfrac{\sqrt{x}}{\sqrt{x}-3}=3\Rightarrow\sqrt{x}=3\sqrt{x}-9\Leftrightarrow2\sqrt{x}=9\Leftrightarrow x=\dfrac{81}{4}\)(tm)
3, Ta có : \(M=P:Q=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{\sqrt{x}+5}{3-\sqrt{x}}=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
