Đáp án+Giải thích các bước giải:
$\frac{1}{(x-1)(x-2)}$ +$\frac{1}{(2-x)(3-x)}$ +$\frac{1}{(1-x)(3-x)}$ DKXD: x $\ne$ 1;2;3$\\$<=> $\frac{1}{(x-1)(x-2)}$ -$\frac{1}{(x-2)(3-x)}$ -$\frac{1}{(x-1)(3-x)}$ $\\$<=>$\frac{3-x}{(x-1)(x-2)(3-x)}$ -$\frac{(x-1)}{(x-1)(x-2)(3-x)}$ -$\frac{x-2}{(x-1)(x-2)(3-x)}$ $\\$<=>$\frac{3-x-x+1-x+2}{(x-1)(x-2)(3-x)}$ <=>$\frac{6-3x}{(x-1)(x-2)(3-x)}$
