Question

Решить неравенство.......

Answer 1

$\frac{1}{x-1}+\frac{1}{2x+1}>\frac{1}{4x+1},x\neq 1;x\neq -\frac{1}{2};x\neq -\frac{1}{4}\\\frac{(2x+1)(4x+1)+(x-1)(4x+1)-(x-1)(2x+1)}{(x-1)(2x+1)(4x+1)}>0\\\frac{8x^2+2x+4x^2+x-2x^2+x+1}{(x-1)(2x+1)(4x+1)}>0\\\frac{10x^2+4x+1}{(x-1)(2x+1)(4x+1)}>0\Leftrightarrow (x-1)(2x+1)(4x+1)>0\\x\in \left ( -\frac{1}{2};-\frac{1}{4} \right )\cup (1;+\infty )$