Question

Помогите пожалуйста решить ​

Answer 1

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

$\displaystyle\bf\\\frac{4\cdot(1-2x)}{(2x-1)(2x+1)} =0\\\\\\\left\{\begin{array}{ccc}1-2x=0\\2x-1\neq 0\\2x+1\neq 0\end{array}\right \\\\\\\left\{\begin{array}{ccc}-2x=-1\\2x\neq 1\\2x\neq -1\end{array}\right \\\\\\\left\{\begin{array}{ccc}x=0,5\\x\neq 0,5\\x\neq -0,5\end{array}\right$

Ответ  :   корней нет