b, 3x-5\(\sqrt{3x+2}\)=-6 (ĐK:x≥\(\dfrac{-2}{3}\))
⇔3x-5\(\sqrt{3x+2}\)+6=0
Đặt \(\sqrt{3x+2}\)=t (t≥0) ⇒ 3x=t\(^2\)-2
⇒ta có pt: t\(^2\)-2-5t+6=0
⇔ t\(^2\)-5t+4=0
⇔t(t-1)+4(t-1)=0
⇔ (t-1)(t+4)=0
⇔t=1 hoặc t=-4(koT/m)⇔t=1(T/m)
⇔3x+2=1⇔3x=-1⇔x=\(\dfrac{-1}{3}\)(TM). Vậy x=\(\dfrac{-1}{3}\)