Question

Помогите! Нужно найти корни уравнения
Нужно расписать действия
$\frac{3x - 5}{ {x}^{2} - 1 } = \frac{3x + 2}{ {x}^{2} + x} - \frac{6x - 5}{ {x}^{2} - x}$
Алгебра 8 класс ​

Answer 1

$\displaystyle\bf\\\frac{3x-5}{x^{2} -1} =\frac{3x+2}{x^{2} +x} -\frac{6x-5}{x^{2} -x} \\\\\\\frac{3x-5}{(x-1)(x+1)} -\frac{3x+2}{x(x+1)} +\frac{6x-5}{x(x-1)} =0\\\\\\\frac{x\cdot(3x-5)-(3x+2)\cdot(x-1)+(6x-5)\cdot(x+1)}{x(x-1)(x+1)} =0\\\\\\\frac{3x^{2} -5x-3x^{2} +3x-2x+2+6x^{2} +6x-5x-5}{x(x-1)(x+1)} =0\\\\\\\frac{6x^{2} -3x-3}{x(x-1)(x+1)} =0\\\\\\\left[\begin{array}{ccc}6x^{2} -3x-3=0\\x\neq 0 \ , \ x\neq 1 \ , \ x\neq -1\end{array}\right \\\\\\6x^{2} -3x-3=0 \ |:3\\\\2x^{2} -x-1=0$

$\displaystyle\bf\\D=(-1)^{2} -4\cdot2\cdot(-1)=1+8=9=3^{2}\\\\\\x_{1} =\frac{1-3}{4} =\frac{-2}{4} =-0,5\\\\\\x_{2} =\frac{1+3}{4} =\frac{4}{4} =1-neyd\\\\\\Otvet \ : \ -0,5$