Question

срочнооооо 30 баллов❗❗❗❗❗❗​

Answer 1

Ответ:

$(-\infty;-\frac{1}{3})\cup(3;10]$

Объяснение:

$\left \{ {{3x^2-8x-3 > 0} \atop {50-5x\geq 0}} \right.\\\\3x^2-8x-3 > 0\\D=(-8)^2-4*3*(-3)=64+36=100=10^2\\x_1=(8+10)/(2*3)=18/6=3\\x_2=(8-10)/(2*3)=-2/6=-1/3\\\\3x^2-8x-3=3(x-3)(x+\frac{1}{3})\\\\50-5x\geq 0\\-5x\geq -50\\x\leq -50:(-5)\\x\leq 10\\\\\\ \left \{ {{3x^2-8x-3 > 0} \atop {50-5x\geq 0}} \right.\; \; \; = > \left \{ {3(x-3)(x+\frac{1}{3}) > 0} \atop {x\leq 10}} \right.$

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/////////////////////////////////////////////////////////////// 10__________

$x\in(-\infty;-\frac{1}{3})\cup(3;10]$