Question

найдите множество решений неравенства 115)

Answer 1

$\displaystyle\bf\\1)\\\\(2x-1)(x+3)\geq 4\\\\2x^{2} +6x-x-3-4\geq 0\\\\2x^{2} +5x-7\geq 0\\\\2(x+3,5)(x-1)\geq 0\\\\(x+3,5)(x-1)\geq 0\\\\\\+ + + + + [-3,5]- - - - - [1]+ + + + + \\\\\\Otvet \ : \ x \in\Big(-\infty \ ; \ -3,5\Big]\cup\Big[1 \ ; \ +\infty\Big)$

$\displaystyle\bf\\2)\\\\(x+2)^{2} < 13-(x-3)^{2} \\\\x^{2} +4x+4 < 13-x^{2} +6x-9\\\\x^{2} +4x+4-4+x^{2} -6x < 0\\\\2x^{2} -2x < 0\\\\x^{2} -x < 0\\\\x\cdot(x-1) < 0\\\\\\+ + + + + (0)- - - - - (1)+ + + + + \\\\\\Otvet \ : \ x\in\Big(0 \ ; \ 1\Big)$

$\displaystyle\bf\\3)\\\\\frac{x^{2} +x}{2}-\frac{8x-1}{3} < -2 \\\\\\\frac{x^{2} +x}{2}\cdot 6-\frac{8x-1}{3}\cdot 6 < -2 \cdot 6\\\\\\(x^{2} +x)\cdot 3-(8x-1)\cdot 2 < -12\\\\3x^{2} +3x-16x+2+12 < 0\\\\3x^{2} -13x+14 < 0\\\\3\cdot\Big(x-2\Big)\cdot\Big(x-2\frac{1}{3} \Big) < 0\\\\\\\Big(x-2\Big)\cdot\Big(x-2\frac{1}{3} \Big) < 0\\\\\\+ + + + + \Big(2\Big)- - - - - \Big(2\frac{1}{3} \Big)+ + + + + \\\\\\Otvet \ :x\in\Big(2 \ ; \ 2\frac{1}{3} \Big)$

$\displaystyle\bf\\4)\\\\\frac{x^{2} -4x}{8} +\frac{x-3}{5} \geq \frac{1-x}{6} \\\\\\\frac{x^{2} -4x}{8} \cdot 120+\frac{x-3}{5} \cdot 120\geq \frac{1-x}{6} \cdot 120\\\\\\(x^{2} -4x)\cdot 15+(x-3)\cdot 24\geq (1-x)\cdot 20\\\\15x^{2} -60x+24x-72\geq 20-20x\\\\15x^{2} -36x-72-20+20x\geq 0\\\\15x^{2} -16x-92\geq 0\\\\15\cdot\Big(x+2\Big)\cdot\Big(x-3\frac{1}{15} \Big)\geq 0\\\\\\+ + + + + \Big[-2\Big]- - - - - \Big[3\frac{1}{15} \Big]+ + + + +$

$\displaystyle\bf\\Otvet \ x\in\Big(-\infty \ ; \ -2\Big]\cup\Big[3\frac{1}{15} \ , \ +\infty\Big)$