Question

N 609. ПОМОГИТЕ РЕШИТЬ.
Система неравенств​

Answer 1

$1)\; \; \left\{\begin{array}{l}x^2-x-2<0\\5-2x<0\end{array}\right\; \; \left\{\begin{array}{l}(x-2)(x+1)<0\\2x>5\end{array}\right\; \; \left\{\begin{array}{l}-1<x<2\\x>2,5\end{array}\right\; \; x\in \varnothing \\\\\\2)\; \; \left\{\begin{array}{l}4+3x-x^2>0\\2-x>0\end{array}\right\; \; \left\{\begin{array}{l}x^2-3x-4<0\\2>x\end{array}\right\; \; \left\{\begin{array}{l}(x-4)(x+1)<0\\x<2\end{array}\right\\\\\\\left\{\begin{array}{l}-1<x<4\\x<2\end{array}\right\; \; \Rightarrow \; \; \; x\in (-1\, ;\, 2)$

$3)\; \; \left\{\begin{array}{l}x^2-8x+15<0\\\frac{1}{2}\, x+1\geq 3\end{array}\right\; \; \left\{\begin{array}{l}(x-3)(x-5)<0\\x+2\geq 6\end{array}\right\; \; \left\{\begin{array}{l}3<x<5\\x\geq 4\end{array}\right\; \; \; \Rightarrow\\\\\\x\in [\; 4\, ;\, 5\, )$

$4)\; \; \left\{\begin{array}{l}-x^2+6x-8<0\\4x-3\leq 0\end{array}\right\; \; \left\{\begin{array}{l}x^2-6x+8>0\\4x\leq 3\end{array}\right\; \; \left\{\begin{array}{l}(x-2)(x-4)>0\\x\leq 0,75\end{array}\right\\\\\\\left\{\begin{array}{l}x\in (-\infty ;2)\cup (4;+\infty )\\x\in (-\infty \, ;\; 0,75\; ]\end{array}\right\; \; \Rightarrow \quad x\in (-\infty \, ;\, 0,75\; ]$