a) $\frac{x - 6}{x - 2}$ + $\frac{x - 1}{x+2}$ = $\frac{3x-4}{4-x^2 }$
⇔ $\frac{(x - 6) (x + 2)}{(x - 2)(x+2)}$ + $\frac{(x - 1)(x-2)}{(x+2)(x-2)}$ = $\frac{4-3x}{x^2-4}$
⇔ $\frac{(x-6)(x+2)}{(x-2)(x+2)}$ + $\frac{(x - 1)(x-2)}{(x+2)(x-2)}$ = $\frac{4-3x}{(x+2)(x-2)}$
⇔ (x-6)(x+2) + (x - 1)(x-2) = 4-3x
⇔ $x^{2}$ - 4x - 12 + $x^{2}$ - 3x + 2 = 4 - 3x
⇔ 2$x^{2}$ - 7x - 10 - 4 + 3x = 0
⇔ 2$x^{2}$ - 4x - 14 = 0
⇔ $x^{2}$ - 2x - 7 = 0
⇔ $x^{2}$ - 2x + 1 - 8 = 0
⇔ $(x - 1)^{2}$ = 8
⇔ x - 1 = ± $\sqrt[]{8}$
⇔ x = 1 + ± $\sqrt[]{8}$
Vậy x ∈ { 1 + ± $\sqrt[]{8}$}
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b) 2(x-1)=3x-2
⇔ 2x - 2 = 3x - 2
⇔ 2x - 3x = -2 + 2
⇔ -x = 0
⇔ x = 0
Vậy x ∈ {0}
