Решите неравенство:
1) (3-5х) (2х²-4х+4) < 0
2) (х²-6х-7) (х²+х+1)≥ 0

[tex]1)\; \; (3-5x)(2x^2-4x+4)\ \textless \ 0\\\\(3-5x)\cdot 2\cdot (x^2-2x+2)\ \textless \ 0\; |:(-2)\\\\(5x-3)(x^2-2x+2)\ \textgreater \ 0\\\\a)\; \; x^2-2x+2=0\; ,\; D=4-8=-4\ \textless \ 0\; \; \to \\\\x^2-2x+2\ \textgreater \ 0\; \; pri\; \; x\in (-\infty ,+\infty )\\\\b)\; \; (5x-3)\underbrace {(x^2-2x+2)}_{\ \textgreater \ 0}\ \textgreater \ 0\; ,\; esli\; \; 5x-3\ \textgreater \ 0\; \; ,\; \; x\ \textgreater \ \frac{3}{5}\\\\Otvet:\; \; x\in (\frac{3}{5},+\infty )\; .[/tex]

[tex]2)\; \; (x^2-6x-7)(x^2+x+1) \geq 0\\\\a)\; \; x^2+x+1=0\; ,\; \; D=1-4=-3\ \textless \ 0\; \; \to \\\\x^2+x+1\ \textgreater \ 0\; \; pri\; \; x\in (-\infty ,+\infty )\\\\b)\; \; (x^2-6x-7)(\underbrace {x^2+x+1}_{\ \textgreater \ 0}) \geq 0\; ,\; esli\; \; x^2-6x-7 \geq 0\\\\x^2-6x-7=0\; ,\; \; x_1=-1\; ,\; \; x_2=7\; \; (teorema\; Vieta)\\\\(x+1)(x-7) \geq 0\\\\+++[-1\, ]---[\, 7\, ]+++\\\\x\in (-\infty ,-1\, ]\cup [\, 7,+\infty )\; \; -\; \; otvet[/tex]