Đáp án:
$\text{ $\dfrac{3}{x-1}$ + $\dfrac{4}{x+1}$ =$\dfrac{3x+2}{1-x²}$ }$
$\text{ ĐKXĐ : x $\neq$ 1 ; x$\neq$ -1 }$
$\text{<=> $\dfrac{3x}{x-1}$ + $\dfrac{4}{x+1}$ = $\dfrac{3x+2}{-(x²-1)}$ }$
$\text{<=> $\dfrac{3}{x-1}$ + $\dfrac{4}{x+1}$ = $\dfrac{-(3x+2)}{x²-1}$}$
$\text{<=> $\dfrac{3(x+1)}{(x-1)(x+1)}$ + $\dfrac{4(x-1)}{(x+1)(x-1)}$ = $\dfrac{-3x-2}{(x-1)(x+1)}$}$
$\text{=> 3(x+1) + 4(x-1) = -3x-2 }$
$\text{<=> 3x+3 + 4x-4 = -3x -2 }$
$\text{<=>3x + 4x+3x = -2 +4 -3 }$
$\text{<=> 10x = -1 }$
$\text{<=> x= -1 : 10}$
$\text{<=> x= $\dfrac{-1}{10}$}$
$\text{Vậy phương trình có tập nghiệm S={$\dfrac{-1}{10}$} }$
