Question

х^2+10х деленное на 2-5х<0

Answer 1

$frac{x^2+10x}{2-5x}<0\f(x)=frac{x^2+10x}{2-5x}\D(f): 2-5xneq0\. xneqfrac{2}{5}\f(x)=0\x^2+10x=0\x(x+10)=0\x=0 x=-10$

Вложение.

$xin (-10;0)cup(frac{2}{5};+infty)$ 

Answer 2

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