Question

зделайте пж даю 50 балов. очень надо прошу срочно. ​

Answer 1

$\displaystyle\bf\\6)\\\\\frac{x+3}{2} -\frac{x-4}{7} < 1\\\\\\\frac{x+3}{2} \cdot 14-\frac{x-4}{7} \cdot 14 < 1\cdot 14\\\\\\(x+3)\cdot 7-(x-4)\cdot 2 < 14\\\\\\7x+21-2x+8 < 14\\\\\\5x+29 < 14\\\\\\5x < -15\\\\\\x < -3\\\\\\Otvet \ : \ x\in(-\infty \ ; \ -3)\\\\7)\\\\\frac{5x-2}{3} +\frac{2x-1}{5} \leq \frac{4-x}{4} \\\\\\\frac{5x-2}{3} \cdot 60+\frac{2x-1}{5} \cdot 60\leq \frac{4-x}{4} \cdot 60\\\\\\(5x-2)\cdot 20+(2x-1)\cdot 12\leq (4-x)\cdot 15\\\\\\100x-40+24x-12\leq 60-15x$

$\displaystyle\bf\\124x-52\leq 60-15x\\\\\\124x+15x\leq 60+52\\\\\\139x\leq 112\\\\\\x\leq \frac{112}{139} \\\\\\Otvet \ : \ x\in\Big(-\infty \ ; \ \frac{112}{139} \Big ]\\\\8)\\\\8(x^{2} -1)-3x(x+2) > 5x^{2} -6x-5\\\\8x^{2} -8-3x^{2} -6x > 5x^{2} -6x-5\\\\5x^{2} -6x-8 > 5x^{2} -6x-5\\\\5x^{2} -6x-5x^{2} +6x > -5+8\\\\0\cdot x > 3\\\\Otvet \ : \ x\in \oslash$

решений нет