Đáp án:
Giải thích các bước giải:
c, $\frac{5-2x}{3}$ + $\frac{(x-1)(x+1)}{3x-1}$ = $\frac{(x+2)(1-3x)}{9x-3}$
ĐKXĐ : x $\neq$ $\frac{1}{3}$
$\frac{5-2x}{3}$ + $\frac{(x-1)(x+1)}{3x-1}$ = $\frac{(x+2)(1-3x)}{9x-3}$
⇔ $\frac{(5-2x)(3x-1)}{3(3x-1)}$ + $\frac{(x-1)(x+1)3}{3(3x-1)}$ = $\frac{(x+2)(1-3x)}{3(3x-1)}$
⇒ (5-2x)(3x-1)+ (x-1)(x+1).3 = (x+2)(1-3x)
⇔ 17x - 6$x^{2}$ - 5 + 3$x^{2}$ - 3 = 2 - 3$x^{2}$ - 5x
⇔ 3$x^{2}$ + 5x + 17x - 6$x^{2}$ + 3$x^{2}$ = 2 + 5 + 3
⇔ 22 x = 10
⇔ x = $\frac{10}{22}$
⇔ x = $\frac{5}{11}$ ( TMĐK )
Vậy x = { $\frac{5}{11}$ }
