Алгебра на картинке, 50 баллов

Решите задачу:

$1)\; \; 2x^2-5x+4 \leq 0\\\\D=25-32=-7\ \textless \ 0\; \; \to \; \; \; 2x^2-5x+4\ \textgreater \ 0\; pri\; x\in (-\infty ,+\infty )\\\\2x^2-5x+4 \leq 0\; \; \Rightarrow \; \; \underline {x\in \varnothing }$

$2)\; \; (3x-8)^2-(4x-6)^2+(5x-2)(5x+2)\ \textgreater \ 96\\\\9x^2-48x+64-(16x^2-48x+36)+25x^2-4-96\ \textgreater \ 0\\\\-7x^2+28+25x^2-100\ \textgreater \ 0\\\\18x^2-72\ \textgreater \ 0\; |:18\\\\x^2-4\ \textgreater \ 0\\\\(x-2)(x+2)\ \textgreater \ 0\\\\+++(-2)---(2)+++\\\\\underline {x\in (-\infty ,-2)\cup (2,+\infty )}$

$3)\; \; \frac{2x+3}{5}-\frac{x^2-4}{8}\leq -1\; |\cdot 40\\\\8(2x+3)-5(x^2-4) \leq -40\\\\16x+24-5x^2+20+40 \leq 0\\\\5x^2-16x-84 \geq 0\\\\D=1936=44^2\; ,\; \; x_1=\frac{16-44}{10}=-2,8\; ,\; \; x_2=\frac{16+44}{10}=6\\\\5\, ( x+2,8)(x-6) \geq 0\\\\+++[-2,8\, ]---[\, 6\, ]+++\\\\\underline {x\in (-\infty ,\, -2,8\, ]\cup [\, 6,+\infty )}$

Алгебра ** картинке, 50 баллов