Question

Решите неравенства,пожалуйста. С 12-го задания.

Answer 1

$12.\ (2 x -5)(x-3) geq 0 \ left { {{2 x -5 geq 0} atop {x-3 geq 0} right. \ left { {{2x geq5 } atop {x geq 3}} right. \ left { {{x geq frac{5}{2} } atop {x geq 3}} right. \ \ 13.\ frac{2x-7}{4-x} geq 0 \ left { {{2x-7 geq 0} atop {4-x neq 0}} right \ left { {{2x geq 7} atop {-x neq -4}} right \ left { {{x geq frac{7}{2} } atop {x neq 4}} right.$

$14. \ left { {{5x+13 leq 0} atop {x+5 geq 1}} right \ left { {{5x leq -13} atop {x geq 1-5}} \ \ left { {{x leq - 2,6 } atop {x geq -4}} right.$

$16. \ left { {{2x-3 leq 5} atop {7-3x leq 1}} right. \ left { {{2x leq 5+3} atop {-3x leq 1-7}} right. \ left { {{2x leq 8} atop {-3x leq -6}} right. \ left { {x leq 4} atop {x geq 2}} right.$


$\17. \ left { {{x^{2} leq 4} atop {x+3 geq 0}} right. \ left { {{ x^{2} leq 4} atop {x geq -3}} right. \ left { {{-2 leq x leq 2} atop {x geq -3}} right.$


$18. \ x^{2} -4x+3 geq 0 \ x_{1} + x_{2}= frac{-b}{a} \ x_{1} * x_{2}= frac{c}{a} \ \ x_{1} + x_{2}= 4 \ x_{1} * x_{2}=3 \ x_{1} geq 3 \ x_{2} geq 1$