tính `a)3/(x^2-2x+1)-6/(x^2-1)+3/(x^2+2x+1)` `b)(x-3)/(4x+4)-(x-1)/(6x-30)`

$\begin{array}{l}a)\quad \dfrac{3}{x^2 – 2x + 1} – \dfrac{6}{x^2-1} + \dfrac{3}{x^2+2x+1}\qquad (ĐK:x \ne \pm 1)\\ = 3\left[\dfrac{1}{(x-1)^2} – \dfrac{2}{(x-1)(x+1)} + \dfrac{1}{(x+1)^2}\right]\\ = 3\cdot\dfrac{(x+1)^2 -2(x-1)(x+1) + (x-1)^2}{(x-1)^2(x+1)^2}\\ = 3\cdot\dfrac{[(x+1) – (x-1)]^2}{(x-1)^2(x+1)^2}\\ = \dfrac{3.2^2}{(x-1)^2(x+1)^2}\\ = \dfrac{12}{(x-1)^2(x+1)^2}\\ b)\quad \dfrac{x-3}{4x+4} – \dfrac{x-1}{6x-30}\qquad (ĐK: x \ne -1;x \ne 5)\\ =\dfrac{x-3}{4(x+1)} – \dfrac{x-1}{6(x-5)}\\ = \dfrac{3(x-3)(x-5) – 2(x-1)(x+1)}{12(x+1)(x-5)}\\ = \dfrac{3(x^2 -8x + 15) – 2(x^2 – 1)}{12(x+1)(x-5)}\\ = \dfrac{3x^2 – 24x + 45 – 2x^2 + 2}{12(x+1)(x-5)}\\ = \dfrac{x^2 -24x + 47}{12(x+1)(x-5)}\\ \end{array}$