$\frac{3}{x^2+x-2}$ + $\frac{2}{x^2-4}$ = $\frac{1}{x^2-3x+2}$

Đáp án + Giải thích các bước giải:

`3/(x^2+x-2) + 2/(x^2-4) = 1/(x^2-3x+2) (x\ne 1; 2;-2)`

`⇔ 3/(x^2-x+2x-2) +2/(x^2-2^2) = 1/(x^2-x-2x+2)`

`⇔ 3/(x(x-1)+2(x-1)) + 2/((x-2)(x+2)) = 1/(x(x-1)-2(x-1))`

`⇔ 3/((x-1)(x+2)) + 2/((x-2)(x+2)) = 1/((x-1)(x-2))`

`⇒ 3(x-2) + 2(x-1) = x+2` 

`⇔ 3x-6+2x-2=x+2`

`⇔ 4x= 10`

`⇔ x = 5/2 (N)`

Vậy `S={5/2}`

$\text{$\dfrac{3}{x² + x - 2}$ + $\dfrac{2}{x² - 4}$ = $\dfrac{1}{x² - 3x + 2}$ Đkxđ: x $\neq$ ±2; x $\neq$ 1}$

$\text{⇔ $\dfrac{3}{x² + 2x - x - 2}$ + $\dfrac{2}{(x - 2)(x + 2)}$ = $\dfrac{1}{x² - x - 2x + 2}$ }$

$\text{⇔ $\dfrac{3}{x(x + 2) - (x + 2)}$ + $\dfrac{2}{(x - 2)(x + 2)}$ = $\dfrac{1}{x(x - 1) - 2(x - 1)}$ }$

$\text{⇔ $\dfrac{3}{(x + 2)(x - 1)}$ + $\dfrac{2}{(x - 2)(x + 2)}$ = $\dfrac{1}{(x - 2)(x - 1)}$ }$

$\text{⇔ $\dfrac{3(x - 2) + 2(x - 1)}{(x - 2)(x + 2)(x - 1)}$ = $\dfrac{1(x + 2)}{(x - 2)(x + 2)(x - 1)}$ }$

$\text{⇒ 3x - 6 + 2x - 2 = x + 2}$

$\text{⇔ 3x + 2x - x = 2 + 6 + 2}$

$\text{⇔ 4x = 10}$

$\text{⇔ x = $\dfrac{5}{2}$ (t/m)}$

$\text{Vậy phương trình có tập nghiệm là S = {$\dfrac{5}{2}$}}$

$\textit{Ha1zzz}$