Первое задание, а б в г.

Решите задачу:

$1)\; (3x+2)(2-3x)\ \textgreater \ (x-2)(2+x)$

$4-9x^2\ \textgreater \ x^2-4\\\\8\ \textgreater \ 10x^2\\\\10x^2-8\ \textless \ 0;5x^2-4\ \textless \ 0\\\\(\sqrt{5}x-2)(\sqrt5x+2)\ \textless \ 0}\\\\+++(-\frac{2}{\sqrt5}})---(\frac{2}{\sqrt5})+++\\\\x\in (-\frac{2}{\sqrt5};\frac{2}{\sqrt5})$

$2)\; \; (x^2+x)(49-x^2)\ \textless \ 0\\\\x(x+1)(7-x)(7+x)\ \textless \ 0\\\\x(x+1)(x-7)(x+7)\ \textgreater \ 0\\\\+++(-7)---(-1)+++(0)---(7)+++\\\\x\in (-\infty -7)\cup (-1,0)\cup (7,+\infty )$

$3)\; \; \frac{(3-2x)(5+3x)}{3x-6} \ \textless \ 0\\\\ \frac{-(2x-3)(3x+5)}{3(x-2)}\ \textless \ 0\; \; \to \; \; \frac{(2x-3)(3x+5)}{3(x-2)} \ \textgreater \ 0\\\\---(-\frac{5}{3})+++(\frac{3}{2})---(2)+++\\\\x\in (-\frac{5}{3},\frac{3}{2})\cup (2,+\infty )$

$4)\; \; \frac{x^3-2x^2+5x-10}{3-2x} \geq 0\\\\ \frac{x^2(x-2)+5(x-2)}{-(2x-3)} \geq 0\; \; \to \; \; \frac{(x-2)(x^2+5)}{2x-3} \leq 0\; \; \; \; (x^2+5\ \textgreater \ 0)\\\\+++(\frac{3}{2})---[\, 2\, ]+++\\\\x\in (\frac{3}{2},2\, ]$