Question

Решить методом интервалов (x^2-5x+4)(x+12)<0

Answer 1

$(x^2-5x+4)(x+12) textless 0\x^2-5x+4=(x-4)(x-1)\x^2-5x+4=0\D=25-16=9=3^2, D textgreater 0\x_{1,2}= frac{5б3}{2} = |{ {{4} atop {1}} right. \(x-4)(x-1)(x+12) textless 0\1)y=(x-4)(x-1)(x+12)\2)y=0,x=4,x=1,x=-12\---(-12)+++1---4+++ textgreater x$
Ответ: $x$∈$(-infty;-12)V(1;4)$