Question

Решите первым методом интервалов:
1) (х+8)(х-3)>0
2) х²-64х<0
Пожалуйста.

Answer 1

$(1)\\\\ (x+8)*(x-3)\ \textgreater \ 0\\\\ \ \ [x-(-8)]*[x-3]\ \textgreater \ 0\\\\ +++++++(-8)--------(3)++++++++++\ \textgreater \ x\\\\ x\in(-\infty;\ -8)\cup(3;\ +\infty)\\\\ -------------------------------\\ (2)\\\\ x^2-64x\ \textless \ 0\\\\ x(x-64)\ \textless \ 0\\\\ (x-0)(x-64)\ \textless \ 0\\\\ ++++++(0)--------(64)+++++++\ \textgreater \ x\\\\ x\in(0;\ 64)$