Question

Решите неравенство ǀx+7ǀ*(x+7)≤16

Answer 1

Ответ:

$|x+7|\cdot (x+7)\leq 16\\\\a)\ \ x\geq -7\ \ \to \ \ \ |x+7|=x+7\ \ \to \\\\(x+7)(x+7)\leq 16\ \ \ \to \ \ (x+7)^2-16\leq 0\ \ \to (x+7-4)(x+7+4)\leq 0\ ,\\\\(x+3)(x+11)\leq 0\ \ \ znaki:\ \ +++[\, -11\ ]---[\, -3\ ]+++\\\\x\in [\, -11\, ;\, -3\ ]\ ,\ \ x\geq -7\ \ \ \Rightarrow \ \ \ \underline {\ x\in [\, -7\, ;\, -3\ ]\ }\\\\b)\ \ x<-7\ \ \ \to \ \ \ |x+7|=-(x+7)\ ,\\\\-(x+7)(x+7)\leq 16\ \ \to \\\\16+(x+7)^2\geq 0\ \ verno\ ,\ esli\x\in R\ ,\ \ tak\ kak\ \ 16>0\ \ i\ \ (x+7)^2\geq 0\ .$

$\underline {\ x\in (-\infty ;-7)\ }\\\\Otvet:\ \ x\in (-\infty \, ;\, -7)\cup [-7\, ;\, -3\ ]=(-\infty \, ;\, -3\ ]\ .$