Question

Решить методом интервалов (x-4)(8-x)(x+5)>0

Answer 1

$(x-4)(8-x)(x+5) textgreater 0, \ -(x-4)(x-8)(x+5) textgreater 0, \ (x-4)(x-8)(x+5) textless 0, \ (x-4)(x-8)(x+5)=0, \ left[begin{array}{c}x-4=0,\x-8=0,\x+5=0;end{array}right. left[begin{array}{c}x=4,\x=8,\x=-5;end{array}right. \ begin{array}{c|ccccccc}x&(-infty;-5)&-5&(-5;4)&4&(4;8)&8&(8;+infty)\(x-4)(8-x)(x+5)&-&0&+&0&-&0&+end{array} \ xin(-infty;-5)cup(4;8).$