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)` 13/11/2021 Bởi Alexandra 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)`
$\begin{array}{l}a)\quad \dfrac{3}{x^2 – 2x + 1} – \dfrac{6}{x^2-1} + \dfrac{3}{x^2+2x+1}\qquad (ĐK:x \ne \pm 1)\\ = 3\left[\dfrac{1}{(x-1)^2} – \dfrac{2}{(x-1)(x+1)} + \dfrac{1}{(x+1)^2}\right]\\ = 3\cdot\dfrac{(x+1)^2 -2(x-1)(x+1) + (x-1)^2}{(x-1)^2(x+1)^2}\\ = 3\cdot\dfrac{[(x+1) – (x-1)]^2}{(x-1)^2(x+1)^2}\\ = \dfrac{3.2^2}{(x-1)^2(x+1)^2}\\ = \dfrac{12}{(x-1)^2(x+1)^2}\\ b)\quad \dfrac{x-3}{4x+4} – \dfrac{x-1}{6x-30}\qquad (ĐK: x \ne -1;x \ne 5)\\ =\dfrac{x-3}{4(x+1)} – \dfrac{x-1}{6(x-5)}\\ = \dfrac{3(x-3)(x-5) – 2(x-1)(x+1)}{12(x+1)(x-5)}\\ = \dfrac{3(x^2 -8x + 15) – 2(x^2 – 1)}{12(x+1)(x-5)}\\ = \dfrac{3x^2 – 24x + 45 – 2x^2 + 2}{12(x+1)(x-5)}\\ = \dfrac{x^2 -24x + 47}{12(x+1)(x-5)}\\ \end{array}$ Bình luận
Đáp án: `a)` ` 3/(x^2 – 2x+1) -6/(x^2-1) + 3/(x^2+2x+1)` ` = 3/((x-1)^2) – 6/((x-1)(x+1)) +3/((x+1)^2)` ` = (3*(x+1)^2 – 6(x^2-1) +3*(x-1)^2)/((x-1)^2 * (x+1)^2)` ` = (3x^2 +6x +3 – 6x^2 +6 + 3x^2 -6x +3)/((x-1)^2 * (x+1)^2)` ` = (12)/((x-1)^2*(x+1)^2) = 12/((x^2-1)^2)` `b)` ` (x-3)/(4x+4) – (x-1)/(6x-30)` ` = (x-3)/(4(x+1)) – (x-1)/(6(x-5))` ` = (3*(x-5)(x-3) – 2*(x+1)(x-1))/(12*(x+1)(x-5))` ` = (3x^2 -24x +45 – 2x^2 +2)/(12*(x+1)(x-5))` ` = (x^2 – 24x+ 47)/(12*(x+1)(x-5))` Bình luận
$\begin{array}{l}a)\quad \dfrac{3}{x^2 – 2x + 1} – \dfrac{6}{x^2-1} + \dfrac{3}{x^2+2x+1}\qquad (ĐK:x \ne \pm 1)\\ = 3\left[\dfrac{1}{(x-1)^2} – \dfrac{2}{(x-1)(x+1)} + \dfrac{1}{(x+1)^2}\right]\\ = 3\cdot\dfrac{(x+1)^2 -2(x-1)(x+1) + (x-1)^2}{(x-1)^2(x+1)^2}\\ = 3\cdot\dfrac{[(x+1) – (x-1)]^2}{(x-1)^2(x+1)^2}\\ = \dfrac{3.2^2}{(x-1)^2(x+1)^2}\\ = \dfrac{12}{(x-1)^2(x+1)^2}\\ b)\quad \dfrac{x-3}{4x+4} – \dfrac{x-1}{6x-30}\qquad (ĐK: x \ne -1;x \ne 5)\\ =\dfrac{x-3}{4(x+1)} – \dfrac{x-1}{6(x-5)}\\ = \dfrac{3(x-3)(x-5) – 2(x-1)(x+1)}{12(x+1)(x-5)}\\ = \dfrac{3(x^2 -8x + 15) – 2(x^2 – 1)}{12(x+1)(x-5)}\\ = \dfrac{3x^2 – 24x + 45 – 2x^2 + 2}{12(x+1)(x-5)}\\ = \dfrac{x^2 -24x + 47}{12(x+1)(x-5)}\\ \end{array}$
Đáp án:
`a)`
` 3/(x^2 – 2x+1) -6/(x^2-1) + 3/(x^2+2x+1)`
` = 3/((x-1)^2) – 6/((x-1)(x+1)) +3/((x+1)^2)`
` = (3*(x+1)^2 – 6(x^2-1) +3*(x-1)^2)/((x-1)^2 * (x+1)^2)`
` = (3x^2 +6x +3 – 6x^2 +6 + 3x^2 -6x +3)/((x-1)^2 * (x+1)^2)`
` = (12)/((x-1)^2*(x+1)^2) = 12/((x^2-1)^2)`
`b)`
` (x-3)/(4x+4) – (x-1)/(6x-30)`
` = (x-3)/(4(x+1)) – (x-1)/(6(x-5))`
` = (3*(x-5)(x-3) – 2*(x+1)(x-1))/(12*(x+1)(x-5))`
` = (3x^2 -24x +45 – 2x^2 +2)/(12*(x+1)(x-5))`
` = (x^2 – 24x+ 47)/(12*(x+1)(x-5))`