Решите уравнения 4 и 6

Решите задачу:

$4)\quad \frac{2}{x-1} - \frac{1}{x+1}\ \textgreater \ 3 \; ,\; \; ODZ:\; x\ne 1,\; \; x\ne -1\\\\ \frac{2(x+1)-(x-1)-3(x-1)(x+1)}{(x-1)(x+1)} \ \textgreater \ 0\\\\ \frac{2x+2-x+1-3x^2+3}{(x-1)(x+1)} \ \textgreater \ 0\\\\ \frac{-3x^2+x+6}{(x-1)(x+1)} \ \textgreater \ 0\; \; \to \; \; \frac{3x^2-x-6}{(x-1)(x+1)} \ \textless \ 0\\\\3x^2-x-6=0\; \; \to \; \; D=1+4\cdot 3\cdot 6=73\; ,\\\\x_1= \frac{1-\sqrt{73}}{6} \approx -1,26\; ,\; \; x_2= \frac{1+\sqrt{73}}{6}\approx 1,59 \\\\ \frac{3(x- \frac{1-\sqrt{73}}{6} )(x- \frac{1+\sqrt{73}}{6} )}{(x-1)(x+1)} \ \textless \ 0$

$+++(-1,26)---(-1)+++(1)---(1,59)+++\\\\x\in (- \frac{1-\sqrt{73}}{6} ,-1)\cup (1, \frac{1+\sqrt{73}}{6} )$

$6)\quad \frac{x-2}{x+2} \geq \frac{2x-3}{4x-1} \; ,\; \; \; \; ODZ:\; x\ne -2\; ,\; x\ne \frac{1}{4}\\\\ \frac{(x-2)(4x-1)-(2x-3)(x+2)}{(x+2)(4x-1)} \geq 0\\\\ \frac{4x^2-9x+2-(2x^2+x-6)}{(x+2)(4x-1)} \geq 0\; \; \to \; \; \frac{2x^2-10x+8}{(x+2)(4x-1)} \geq 0\\\\ \frac{2(x-1)(x-4)}{(x+2)(4x-1)} \geq 0\\\\+++(-2)---(\frac{1}{4})+++[\; 1\, ]---[\, 4\, ]+++ \\\\x\in (-\infty ,-2)\cup (\frac{1}{4},1\, ]\cup [\, 4,+\infty )$