\(x-3\sqrt{x}+2=0\)
\(\sqrt{x^2}-3\sqrt{x}+2=0\)
\(\sqrt{x}\left(\sqrt{x}-3\right)=-2=1.\left(-2\right)\)
=> \(\sqrt{x}=1\)=> x =1;
Vậy x = 1
ta có \(x-3\sqrt{x}+2=0\)
=> \(x-3\sqrt{x}=-2\)
=> \(\sqrt{x}\left(\sqrt{x}-3\right)=-2=-1\cdot2=-2\cdot1\)
Mà \(\sqrt{x}>=0\) => \(\sqrt{x}=1hoặc\sqrt{x}=2\)
=> x=1 hoặc x=4
Vậy ...