Đáp án:
a) ĐKXĐ: $x>1_{}$
b) $\frac{x+5}{x+1}$
Giải thích các bước giải:
$\frac{x}{x-1}$ + $\frac{3}{x-1}$ - $\frac{6x-4}{x^2-1}$
= $\frac{x}{x-1}$ + $\frac{3}{x-1}$ - $\frac{6x-4}{(x-1)(x+1)}$ $(*)_{}$
a) ĐKXĐ: $x-1>0_{}$
⇔ $x>1_{}$
b) $(*)_{}$ $\frac{x(x+1)}{(x-1)(x+1)}$ + $\frac{3(x+1)}{(x-1)(x+1)}$ - $\frac{6x-4}{(x-1)(x+1)}$
= $\frac{x^2+x}{(x-1)(x+1)}$ + $\frac{3x+3}{(x-1)(x+1)}$ - $\frac{6x-4}{(x-1)(x+1)}$
= $\frac{x^2+x+3x+3-6x+4}{(x-1)(x+1)}$
= $\frac{x^2-2x+7}{(x-1)(x+1)}$
= $\frac{x^2-2x+1+6}{(x-1)(x+1)}$
= $\frac{(x-1)^2+6}{(x-1)(x+1)}$
= $\frac{x-1+6}{x+1}$
= $\frac{x+5}{x+1}$
