Question

БУДУ ОЧЕНЬ БЛАГОДАРНА ЕСЛИ КТО ПОМОЖЕТ!!!!!! ОЧЕНЬ НАДОО!!!!!


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Answer 1

Ответ:

$1)\displaystyle x\in(-2;+\infty);\\2)x\in[-2;+\infty);\\3)x\in\bigg[-\frac{2}{3} ;+\infty\Big);\\4)x\in\varnothing;\\5)x\in[-0,35;+\infty);\\6)x\in(-2;0);\\7)x\in(-\infty;7)$

Объяснение:

$\displaystyle 1)-6x < 12;\\6x > -12;\\x > -2;\\otvet:x\in(-2;+\infty);\\2)\frac{2x}{4} \geq -1;\\2x\geq -4;\\x\geq -2;\\----[-2]//////\Rightarrow_X\\otvet:x\in[-2;+\infty)\\3) 1+4y\leq 7y+3;\\-3y\leq 2;\\3y\geq -2;\\y\geq -\frac{2}{3} ;\\otvet:x\in\bigg[-\frac{2}{3} ;+\infty\Big)$

$\displaystyle 4) \left \{ {{-2x > 3} \atop {-6x < 4;}} \right.\Leftrightarrow \left \{ {{-x > \frac{3}{2}, } \atop {-x < \frac{4}{6} ;}}\right.\Leftrightarrow\left \{ {{x < -1,5,} \atop {x > -\frac{2}{3}}} \right. \\otvet:x\in\varnothing\\5)\left \{ {{2x+4\geq -3(6x+1),} \atop {x^{2} +4\geq (x-2)(x+2)+x;}} \right. \Leftrightarrow \left \{ {{2x+4\geq -18x-3,} \atop {x^{2} +4\geq x^{2} -4+x;}} \right. \Leftrightarrow \left \{ {{20x\geq -7,} \atop {x\geq 4+4;}} \Leftrightarrow$

$\Leftrightarrow \displaystyle \left \{ {{x\geq -0,35,} \atop {x\geq 8}} \right. \\otvet:x\in[-0,35;+\infty)\\6)-4 < \frac{5x+2}{2} < 1;\\-8 < 5x+2 < 2;\\\left \{ {{5x+2 > -8,} \atop {5x+2 < 2;}} \right. \Leftrightarrow\left \{ {{5x < -10,} \atop {5x < 0;}} \right. \Leftrightarrow\left \{ {{x > -2,} \atop {x < 0}} \right. \\otvet: x\in(-2;0)$

$\displaystyle 7)\left \{ {{\frac{2x+1}{6}+\frac{x+2^{(2}}{3} -\frac{x-8^{(3}}{2} > x-1^{(6},} \atop {-2x+2 > 0,5+0,5x;}} \right. \Leftrightarrow\left \{ {{2x+1+2x+4-3x+24 > 6x-6,} \atop {-2x-0,5x > 0,5-2;}} \right. \Leftrightarrow\left \{ {{2x+2x-3x-6x > -6-1-4-24,} \atop {-2,5x > 0,5-2;}} \right. \Leftrightarrow\left \{ {{-5x > -35,} \atop {-2,5x > -1,5;}} \right. \Leftrightarrow\left \{ {{5x < 35,} \atop {2,5x < 1,5;}} \Leftrightarrow$$\Leftrightarrow\displaystyle \left \{ {{x < 7,} \atop {x < 0,6}} \right. \\otvet:x\in(-\infty;7)$