Question

Срочно!!!! даю максимум баллов.​

Answer 1

$\displaystyle\bf\\\left \{ {{1+\dfrac{2x+3}{3} < \dfrac{x+1}{4} } \atop {6(2x-1)+7 < 5(x-4)}} \right. \\\\\\\left \{ {{1\cdot 12+\dfrac{2x+3}{3} \cdot 12 < \dfrac{x+1}{4}\cdot 12 } \atop {12x-6+7 < 5x-20}} \right. \\\\\\\left \{ {{12+(2x+3)\cdot 4 < (x+1)\cdot 3} \atop {12x+1 < 5x-20}} \right.\\\\\\\left \{ {{12+8x+12 < 3x+3} \atop {12x-5x < -20-1}} \right. \\\\\\\left \{ {{8x+24 < 3x+3} \atop {7x < -21}} \right. \\\\\\\left \{ {{8x-3x < 3-24} \atop {7x < -21}} \right.$

$\displaystyle\bf\\\left \{ {{5x < -21} \atop {x < -3}} \right. \\\\\\\left \{ {{x < -4,2} \atop {x < -3}} \right. \ \ \ \Rightarrow \ \ \ x < -4,2\\\\\\Otvet \ : \ x\in\Big(-\infty \ ; \ -4,2\Big)$