Предмет: Алгебра, автор: misivhristina9

Очень нужно!!!!!!!!!!!!!!

Приложения:

Ответы

Автор ответа: emilkadyrov

Ответ:

ответ получилось у меня -6/x8

я не смогла 2,3, 4.

Приложения:

misivhristina9: Спасибо

Автор ответа: boberon

$\dfrac{2x-3}{x^2+4x+4} - \dfrac{x+1}{x^2+2x} = \dfrac{5}{x}\\\\\dfrac{2x-3}{(x+2)(x+2)} - \dfrac{x+1}{x(x+2)} - \dfrac{5}{x} = 0\\\\\dfrac{(2x-3)x - (x+1)(x+2) - 5(x+2)(x+2)}{x(x+2)(x+2)} = 0\\\\(2x-3)x - (x+1)(x+2) - 5(x+2)(x+2) = 0, \ x \neq 0, x \neq -2\\\\2x^2 - 3x - x^2 - 3x - 2 - 5x^2 - 20x - 20 = 0\\\\-4x^2 - 26x - 22 = 0\\\\2x^2 + 13x + 11 = 0\\\\D = 13^2 - 4\cdot2\cdot11 = 81 = 9^2\\\\x_1 = \dfrac{-13-9}{4} = -\dfrac{22}{4} = -\dfrac{11}{2}\\\\x_2 = \dfrac{-13+9}{4} = -1$

Ответ: x₁ = –11/2; x₂ = –1.

$\dfrac{6}{x^2-9} - \dfrac{4}{x^2+6x+9} = \dfrac{1}{x-3}\\\\\dfrac{6}{(x-3)(x+3)} - \dfrac{4}{(x+3)(x+3)} - \dfrac{1}{x-3} = 0\\\\\dfrac{6(x+3) - 4(x-3) - (x+3)(x+3)}{(x-3)(x+3)(x+3)} = 0\\\\6(x+3) - 4(x-3) - (x+3)(x+3) = 0, \ x \neq \pm 3 \ (!!!)\\\\6x + 18 - 4x + 12 -x^2 - 6x - 9 = 0\\\\-x^2 - 4x + 21 = 0\\\\D = (-4)^2 + 4\cdot21 = 100 = 10^2\\\\x_1 = \dfrac{4 - 10}{-2} = 3, \ x \neq 3 \ (!!!)\\\\x_2 = \dfrac{4+10}{-2} = -7$

Ответ: x = –7.

$\dfrac{6}{x^2-36} - \dfrac{3}{x^2+6x} = \dfrac{x+12}{x^2-6x}\\\\\dfrac{6}{(x-6)(x+6)} - \dfrac{3}{x(x+6)} - \dfrac{x+12}{x(x-6)} = 0\\\\\dfrac{6x - 3(x-6) - (x+12)(x+6)}{x(x-6)(x+6)} = 0\\\\6x - 3(x-6) - (x+12)(x+6) = 0, \ x \neq 0, x \neq \pm 6 \ (!!!)\\\\6x - 3x + 18 - x^2 - 18x - 72 = 0\\\\-x^2 - 15x - 54 = 0\\\\D = (-15)^2 - 4\cdot54 = 9 = 3^2\\\\x_1 = \dfrac{15-3}{-2} = -6, \ x \neq -6 \ (!!!)\\\\x_2 = \dfrac{15+3}{-2} = -9$

Ответ: x = –9.

$\dfrac{3x+2}{x+1} + \dfrac{x+4}{x-3} = \dfrac{3x^2+1}{x^2-2x-3}\\\\\dfrac{3x+2}{x+1} + \dfrac{x+4}{x-3} - \dfrac{3x^2+1}{(x+1)(x-3)} = 0\\\\\dfrac{(3x+2)(x-3) + (x+4)(x+1) - (3x^2+1)}{(x+1)(x-3)} = 0\\\\(3x+2)(x-3) + (x+4)(x+1) - (3x^2+1) = 0, \ x \neq -1, x \neq 3 \ (!!!)\\\\3x^2 - 7x - 6 + x^2 + 5x + 4 - 3x^2 - 1 = 0\\\\x^2 - 2x - 3 = 0\\\\(x+1)(x-3) = 0\\\\x_1 = -1, \ x \neq -1 \ (!!!)\\\\x_2 = 3, \ x \neq 3 \ (!!!)$