a) $\frac{4x-2}{6}$ + $\frac{3-2x}{2}$ = 2 - $\frac{x+7}{3}$
⇔ $\frac{4x-2}{6}$ + $\frac{9-6x}{6}$ = $\frac{12}{6}$ - $\frac{2x+14}{6}$
⇒ 4x - 2 + 9 - 6x = 12 - 2x -14
⇔ 0x = -9
Vậy pt vô nghiệm
b) $\frac{5}{x+2}$ - $\frac{3}{x-2}$ = $\frac{2(x-9)}{x^2-2^2}$
⇔$\frac{5x-10}{(x+2)(x-2)}$ - $\frac{3x+6}{(x+2)(x-2)}$ = $\frac{2x-18}{(x+2)(x-2)}$
⇒ 5x - 10 - 3x - 6 = 2x - 18
⇔ 0x = -2
Vậy pt vô nghiệm
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