Question

Решите неравенство 2х^2-8х_>0
Решите систему
Х+4_>0
Х^2-х-56<0

Answer 1

$1); ; 2x^2-8x textgreater 0\\2x(x-4) textgreater 0; ; ; ; ; +++(0)---(4)+++\\xin (-infty ,0)cup (4,+infty )$

$2); ; left { {{x+4 textgreater 0} atop {x^2-x-56 textless 0}} right. ; left { {{x textgreater -4} atop {(x+7)(x-8) textless 0}} right. ; left { {{xin (-4,+infty )} atop {xin (-7,8)}} right. ; Rightarrow ; ; xin (-4,8)\\x^2-x-56=0; to x_1=-7,; x_2=8; ; (teor.; Vieta)\\+++(-7)---(8)+++$