1.
b) 3(2x - 1) + 8 = x - 5
6x - 3 + 8 = x - 5
5x = -10
x = -2
Vậy x = -2
d) x(3x + 2) -2x² = x(x - 8) + 4
3x² + 2x - 2x² = x² -8x + 4
10x = 4
x = $\frac{2}{5}$
Vậy x = $\frac{2}{5}$
h) (x - 1)(2x + 5)(3x - 6) = 0
TH1: x - 1 = 0
x = 1
TH2: 2x + 5 = 0
x = $\frac{-5}{2}$
TH3: 3x - 6 = 0
x = 2
Vậy x ∈ { 1; $\frac{-5}{2}$ ; 2 }
2 b)
$\frac{2x-1}{x-3)}$ + $\frac{1}{2}$ = $\frac{x}{x-3}$ ( x khác 3)
$\frac{2(2x-1)}{2(x-3)}$ + $\frac{x-3}{2(x-3)}$ = $\frac{2x}{x-3}$
2(2x - 1) + x - 3 = 2x
4x - 2 + x - 3 = 2x
3x = 5
x = $\frac{5}{3}$ (tm)
Vậy x = $\frac{5}{3}$
