$\frac{4}{x+6}$ + $\frac{1}{x-3}$ = $\frac{3x+8}{x^2+7x+6}$

$\text{$\dfrac{4}{x + 6}$ + $\dfrac{1}{x - 3}$ = $\dfrac{3x + 8}{x² + 7x + 6}$ Đkxđ: x $\neq$ -1; x $\neq$ 3; x $\neq$ -6}$

$\text{⇔ $\dfrac{4(x - 3) + 1(x + 6)}{(x - 3)(x + 6)}$ = $\dfrac{3x + 8}{x² + x + 6x + 6}$ }$

$\text{⇔ $\dfrac{4x - 12 + x + 6}{(x - 3)(x + 6)}$ = $\dfrac{3x + 8}{x(x + 1) + 6(x + 1)}$ }$

$\text{⇔ $\dfrac{5x - 6}{(x - 3)(x + 6)}$ = $\dfrac{3x + 8}{(x + 6)(x + 1)}$ }$

$\text{⇔ $\dfrac{(5x - 6)(x + 1)}{(x - 3)(x + 6)(x + 1)}$ = $\dfrac{(3x + 8)(x - 3)}{(x + 6)(x + 1)(x - 3)}$ }$

$\text{⇒ 5x² + 5x - 6x - 6 = 3x² - 9x + 8x - 24}$

$\text{⇔ 5x² - x - 6 = 3x² - x - 24}$

$\text{⇔ 5x² - 3x² - x + x = -24 + 6}$

$\text{⇔ 2x² = -18}$

$\text{⇔ x² = -9}$

$\text{Mà x² ≥ 0 với ∀ x ∈ đkxđ}$

$\text{⇒ x ∈ ∅}$

$\text{Vậy phương trình vô nghiệm}$

$\textit{Ha1zzz}$

Đáp án:

 `S={∅}`

Giải thích các bước giải:

 `4/(x+6)+1/(x-3)=(3x+8)/(x^2+7x+6)(ĐKXD: x \ne -6;3;-1)`

`=>(4(x-3)+x+6)/((x+6)(x-3))=(3x+8)/((x+6)(x+1))`

`=>((5x-6)(x+1))/((x+6)(x-3)(x+1))=((3x+8)(x-3))/((x+6)(x-3)(x+1))`

`=>(5x-6)(x+1)=(3x+8)(x-3)`

`=>(5x^2-x-6)=(3x^2-x-24)`

`=>2x^2+18=0`

`=>x^2+9=0` ( vô lí )

`=>x in ∅`

`@nguyen``nam500#hoidap247`