Question

Розв'яжіть рівняння:
$\frac{4}{1 + x} - \frac{x + 1}{x - 1} = \frac{3 - {x}^{2} }{ {x}^{2} - 1}$
Помогите пожалуйста!!!! Даю 50 балов​

Answer 1

$\displaystyle\bf\\\frac{4}{1+x} -\frac{x+1}{x-1} =\frac{3-x^{2} }{x^{2} -1} \\\\\\\frac{4}{x+1} -\frac{x+1}{x-1} -\frac{3-x^{2} }{(x -1)\cdot(x+1)} =0\\\\\\\frac{4\cdot(x-1)-(x+1)\cdot(x+1)-3+x^{2} }{(x-1)\cdot(x+1)} =0\\\\\\\frac{4x-4-x^{2} -2x-1-3+x^{2} }{(x-1)\cdot(x+1)} =0\\\\\\\frac{2x-8}{(x-1)\cdot(x+1)} =0$

$\displaystyle\bf\\\left\{\begin{array}{ccc}2x-8=0\\x-1\neq 0\\x+1\neq 0\end{array}\right\\\\\\\left\{\begin{array}{ccc}x=4\\x\neq 1\\x\neq -1\end{array}\right\\\\\\Otvet \ : \ 4$