\(\left(3x-1\right)\left(2x-3\right)\left(2x-3\right)\left(x+5\right)=0\)
Th1 : \(3x-1=0=>x=\frac{1}{3}\)
Th2 : \(2x-3=0=>x=\frac{3}{2}\)
TH3 : \(x+5=0=>x=-5\)
Mik tl mà chẳng có ai T kì quá z
(3x - 1)(2x - 3)(2x - 3)(x + 5) = 0
=> (3x - 1)(2x - 3)2(x + 5) = 0
=> 3x - 1 = 0 => x = 1/3
hoặc 2x - 3 = 0 => x = 3/2
hoặc x + 5 = 0 => x = -5
Vậy x = 1/3 , x = 3/2 , x = -5