\(3x-7\sqrt{x}+4=0\) (ĐK: x≥0)
⇔ \(3\left(\sqrt{x}\right)^2-7\sqrt{x}+4=0\)
⇔ \(3\left(\sqrt{x}\right)^2-3\sqrt{x}-4\sqrt{x}+4=0\)
⇔ \(\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\)=0
⇔ [ \(3\sqrt{x}-4\)=0 ⇔ [x=\(\dfrac{16}{9}\)
[ \(\sqrt{x}-1=0\) [x=1
Vậy x=\(\dfrac{16}{9}\) , x=1

