1) \(đk:x\ge0,x\ne4\)
\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b) \(x=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=3+\sqrt{2}+3-\sqrt{2}=6\)
\(P=\dfrac{3\sqrt{6}}{\sqrt{6}+2}=\dfrac{3\sqrt{6}\left(\sqrt{6}-2\right)}{6-4}=\dfrac{18-6\sqrt{6}}{2}=9-3\sqrt{6}\)
c) \(P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}>2\Leftrightarrow3\sqrt{x}>2\sqrt{x}+4\Leftrightarrow\sqrt{x}>4\)
Kết hợp đk
\(\Leftrightarrow x>16\)
4) \(P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}=3-\dfrac{6}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Do \(x\ge0,x\ne4\)
\(\Rightarrow x\in\left\{0;1;16\right\}\)
hép mi
somebody help me plz =))