Question

найдите число целых решений неравенства

Answer 1

$\frac{6^{x+1}-3^{2x-1}*2^{4x-5}}{ \sqrt{x+2} } \geq 0\\ x+2\ \textgreater \ 0, \ x\ \textgreater \ -2\\ 6^{x+1}-3^{2x-1}*2^{4x-5} \geq 0\\ 6*6^x- \frac{3^{2x}}{3} * \frac{2^{4x}}{2^5} \geq 0/:6\\ 6^x- \frac{9^x*16^x}{576} \geq 0\\ 6^x \geq \frac{144^x}{24^2}/:144^x\\ (\frac{6}{144}) ^x \geq ( \frac{1}{24}) ^2\\ ( \frac{1}{24} )^x \geq ( \frac{1}{24} )^2\\ x \leq 2\\ \left \{ {{x\ \textgreater \ -2} \atop {x \leq 2}} \right. \\ x\in(-2;2]\\ x=-1;\ x=0; \ x=1; \ x=2$
Ответ: 4