Question

Решите систему неравенств. алгебра 11класс. пожалуйста

Answer 1

$\left \{ {{x^2+x-6 \geq 0} \atop {log_4^2x-log_4x-6\ \textless \ 0}} \right. \\\\1)\; \; x^2+x-6 \geq 0\; \; \; \Rightarrow \; \; \; x_1=-3\; ,\; \; x_2=2\; \; \; (teorema\; Vieta)\\\\+++[-3\, ]---[\, 2\, ]+++\\\\x\in (-\infty ,-3\, ]\cup [\, 2,+\infty )\\\\2)\; \; log_4^2x-log_4x-6\ \textless \ 0$

$t=log_4x\; ,\; \; \; t^2-t-6\ \textless \ 0\; ,\; \; t_1=-2,\; t_2=3\\\\+++(-2)---(3)+++\\\\-2\ \textless \ t\ \textless \ 3\\\\ \left \{ {{log_4x\ \textgreater \ -2} \atop {log_4x\ \textless \ 3}} \right. \; \left \{ {{log_4x\ \textgreater \ log_44^{-2}} \atop {log_4x\ \textless \ log_44^3}} \right. \; \left \{ {{x\ \textgreater \ \frac{1}{16}} \atop {x\ \textless \ 64}} \right. \; x\in (\frac{1}{16}\, ,\, 64)$

$3)\; \; \; \left \{ {{x\in (-\infty ,-3\, ]\cup [\, 2,+\infty )} \atop {x\in (\frac{1}{16},64)}} \right. \; \; \Rightarrow \; \; \; \underline {x\in [\, 2,64)}\\\\Otvet: B\; .$