Question

Решите пожалуйста)) $5x-17\sqrt{x+5} +31 \ \textless \ 0$

Answer 1

ОДЗ: $\displaystyle \tt x\geq -5$

$\displaystyle \tt 5x-17\sqrt{x+5}+31<0\\\displaystyle \tt -17\sqrt{x+5}<-5x-31\\\displaystyle \tt 17\sqrt{x+5}>5x+31\\\\ \displaystyle \tt 17\sqrt{x+5}>5x+31, \: \: \: \: 5x+31\geq 0\\ \displaystyle \tt 17\sqrt{x+5}>5x+31, \: \: \: \: 5x+31<0\\\\ \displaystyle \tt x\in\bigg(-\frac{121}{25}; \: 4\bigg), \: \: \: \: x\geq -\frac{31}{5}\\\displaystyle \tt x\in R, \: \: \: \: x<-\frac{31}{5}$

$\displaystyle \tt x\in\bigg(-\frac{121}{25}; \: 4\bigg)\\\displaystyle \tt x\in\bigg(-\infty; \: -\frac{31}{5}\bigg)\\\\\\ x\in\bigg(-\infty; \: -\frac{31}{5}\bigg)\cup \bigg(-\frac{121}{25}; \: 4\bigg), \: \: \: \: x\geq -5\\\\\\ \displaystyle \tt x\in\bigg(-\frac{121}{25}; \: 4\bigg)$

ОТВЕТ:

$\displaystyle \tt x\in\bigg(-\frac{121}{25}; \: 4\bigg)$

Answer 2

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