a) $x^{3}$ + 5x = 0
(=) x(x² + 5) = 0
TH1: x = 0
TH2: x² + 5 = 0
(=) x² = -5 (vô lí)
Vậy x = 0
b) 2x(x - 3) + (x - 3) = 0
(=) (x-3)(2x+1) = 0
TH1: x - 3 =0
(=) x = 3
TH2: 2x + 1 = 0
(=) x = $\frac{-1}{2}$
Vậy x = 3 hoặc x = $\frac{-1}{2}$
c) (x+2)(x-1) - x+1 = 0
(=) (x+2)(x-1) - (x-1) = 0
(=) (x-1)(x+2-1) = 0
(=) (x-1)(x+1) = 0
TH1: x - 1 =0
(=) x= 1
TH2: x + 1 = 0
(=) x= -1
Vậy x = ±1
d) x+ 3 = (x+3)²
(=) x + 3 = x² + 6x + 9
(=) x + 3 - x² - 6x - 9 = 0
(=) -x² - 5x - 6=0
(=) -1.(-x²) -1. (-5x) -1. (-6)= 0
(=) x² + 5x + 6=0
(=) x² + 3x + 2x + 6 = 0
(=) x(x+3) + 2(x+3)=0
(=) (x+3)(x+2)=0
TH1: x + 3 =0
(=) x=-3
TH2: x + 2 = 0
(=) x=-2
Vậy x=-3 hoặc x=-2
f) x(x+1) + 2(x+1) = 0
(=) (x+1)(x+2) = 0
TH1: x+1=0
(=) x = -1
TH2: x+2=0
(=) x = -2
Vậy x=-1 hoặc x=-2
g) x(x+1) - x -1 = 0
(=) x(x+1) - (x +1) = 0
(=) (x+1)(x-1) = 0
TH1: x+1=0
(=) x=-1
TH2: x-1=0
(=) x=1
Vậy x= ±1
h) 2x + 1= (2x+1)²
(=) 2x + 1 = 4x² + 4x + 1
(=) 2x + 1 - 4x² - 4x - 1=0
(=) -4x² -2x =0
(=) -2x(1+2x)=0
(=) x(1+2x)=0
TH1: x=0
TH2: 1+2x =0
(=) x= $\frac{-1}{2}$
Vậy x=0 hoặc x= $\frac{-1}{2}$
