Question

первое сократить второе вычислить пж пж
8 класс

Answer 1

Ответ:

Объяснение:

Каждый квадратный трехчлен раскладываем на множители, находя корни $x_1$ и $x_2$ по теореме Виета или через дискриминант, следующим образом: $ax^2+bx+c=a(x-x_1)(x-x_2).$

6.

$1)\,\,\displaystyle\frac{{4{x^2} - 81}}{{2{x^2} - 5x - 18}} = \displaystyle\frac{{(2x - 9)(2x + 9)}}{{(2x - 9)(x + 2)}} = \displaystyle\frac{{2x + 9}}{{x + 2}};$

$2)\,\,\displaystyle\frac{{2{x^2} + 6x - 20}}{{{x^3} - 8}} = \displaystyle\frac{{2(x + 5)(x - 2)}}{{(x - 2)({x^2} + 2x + 4)}} = \displaystyle\frac{{2(x + 5)}}{{{x^2} + 2x + 4}};$

$3)\,\,\displaystyle\frac{{2{x^2} - 12x + 18}}{{2{x^2} - x - 15}} = \displaystyle\frac{{2{{(x - 3)}^2}}}{{(2x + 5)(x - 3)}} = \displaystyle\frac{{2(x - 3)}}{{2x + 5}};$

$\\\\4)\,\,\displaystyle\frac{{4{x^2} - 11x - 3}}{{ - 3{x^2} + 10x - 3}} = \displaystyle\frac{{(x - 3)(4x + 1)}}{{ - (x - 3)(3x - 1)}} = \displaystyle\frac{{4x + 1}}{{1 - 3x}}.$

7.

$1)\,\,\displaystyle\frac{{x - 1}}{{{x^2} + 2x - 3}} + \displaystyle\frac{{x + 1}}{{{x^2} + 4x + 3}} = \displaystyle\frac{{x - 1}}{{(x + 3)(x - 1)}} + \displaystyle\frac{{x + 1}}{{(x + 3)(x + 1)}} =\\\\= \displaystyle\frac{1}{{x + 3}} + \displaystyle\frac{1}{{x + 3}} = \displaystyle\frac{2}{{x + 3}};$

$2)\,\,\displaystyle\frac{{2{x^2} - 7}}{{{x^2} - 3x - 4}} - \displaystyle\frac{{x + 1}}{{x - 4}} = \displaystyle\frac{{2{x^2} - 7}}{{(x - 4)(x + 1)}} - \displaystyle\frac{{x + 1}}{{x - 4}} = \displaystyle\frac{{2{x^2} - 7 - {{(x + 1)}^2}}}{{(x - 4)(x + 1)}} =\\\\= \displaystyle\frac{{2{x^2} - 7 - {x^2} - 2x - 1}}{{(x - 4)(x + 1)}} = \displaystyle\frac{{{x^2} - 2x - 8}}{{(x - 4)(x + 1)}} = \displaystyle\frac{{(x - 4)(x + 2)}}{{(x - 4)(x + 1)}} = \displaystyle\frac{{x + 2}}{{x + 1}};$

$3)\,\,\displaystyle\frac{{{x^2} - x - 20}}{{2 - x}} \cdot \displaystyle\frac{{2x - {x^2}}}{{x + 4}} = \displaystyle\frac{{(x - 5)(x + 4)}}{{2 - x}} \cdot \displaystyle\frac{{x(2 - x)}}{{x + 4}} = x(x - 5);$

$4)\,\,\displaystyle\frac{{x + 5}}{{2x - 6}}:\displaystyle\frac{{{x^2} + 11x + 30}}{{x - 3}} = \displaystyle\frac{{x + 5}}{{2(x - 3)}} \cdot \displaystyle\frac{{x - 3}}{{(x + 5)(x + 6)}} = \displaystyle\frac{1}{{2(x + 6)}}.$