c, 1 + $\frac{x}{3-x}$ = $\frac{5x}{(x+2)(3-x)}$ + $\frac{2}{x+2}$ (ĐKXĐ: x$\neq$-2; x$\neq$3)
⇔$\frac{(3-x)(x+2)}{(3-x)(x+2)}$ + $\frac{x(x+2)}{(3-x)(x+2)}$ = $\frac{5x}{(x+2)(3-x)}$ + $\frac{2(3-x)}{(x+2)(3-x)}$
⇔ $\frac{3x+6-x^2-2x}{(3-x)(x+2)}$ + $\frac{x^2+2x}{(3-x)(x+2)}$ = $\frac{5x}{(x+2)(3-x)}$ + $\frac{6-2x}{(x+2)(3-x)}$
⇒ 3x + 6 - x² - 2x + x² + 2x = 5x + 6 - 2x
⇔ 3x - x² - 2x + x² + 2x - 5x + 2x = 6 - 6
⇔ 0x = 0 (luôn đúng)
Kết hợp với ĐKXĐ, ta có x ∈ R; x$\neq$-2; x$\neq$3
d, |-2x| - 3 - 5x + x = 0
⇔ |-2x| -3 - 4x = 0
⇔ |-2x| = 4x + 3
⇔ -2x = 4x + 3 khi -2x $\geq$ 0
hoặc -2x = -4x - 3 khi -2x < 0
⇔ -2x - 4x = 3 khi x $\leq$ 0
hoặc -2x + 4x = -3 khi x > 0
⇔ -6x = 3 khi x $\leq$ 0
hoặc 2x = -3 khi x > 0
⇔ x = $\frac{-1}{2}$ (tmđk)
hoặc x = $\frac{-3}{2}$ (ktmđk)
Vậy x = $\frac{-1}{2}$
