Решите неравенство: (2x+1)-² < (49/16)^-1 Те кто не понял: (2х+1) в минус второй степени...

Решите задачу:

$(2x+1)^{-2}\ \textless \ (\frac{49}{16})^{-1}\\\\\frac{1}{(2x+1)^2}\ \textless \ \frac{16}{49}\\\\(\frac{1}{2x+1})^2-(\frac{4}{7})^2\ \textless \ 0\\\\(\frac{1}{2x+1}-\frac{4}{7})(\frac{1}{2x+1}+\frac{4}{7})\ \textless \ 0\\\\\frac{7-4(2x+1)}{7(2x+1)}\cdot \frac{7+4(2x+1)}{7(2x+1)}\ \textless \ 0\\\\\frac{(-8x+3)(8x+11)}{7^2(2x+1)^2}\ \textless \ 0\\\\7^2(2x+1)^2\ \textgreater \ 0\; pri\; x\ne -\frac{1}{2}\; \; \Rightarrow \; \; -(8x-3)(8x+11)\ \textless \ 0\\\\(8x-3)(8x+11)\ \textgreater \ 0$

$+++(-\frac{11}{8})---(\frac{3}{8})+++\\\\x\in (-\infty ,-\frac{11}{8})\cup (\frac{3}{8},+\infty )$