Question

Помогите решить неравенства:
$6^{x} +( \frac{1}{6} )^{x}$>2
$2^{x^{2}} \leq 64$ · $2^{x}$

Answer 1

$1)\; 6^{x}+(\frac{1}{6})^{x}\ \textgreater \ 2\\\\6^{x}+\frac{1}{6^{x}}-2\ \textgreater \ 0\, |\cdot 6^{x}\ \textgreater \ 0\\\\(6^{x})^2-2\cdot 6^{x}+1\ \textgreater \ 0\\\\t=6^{x},\; \; t^2-2t+1\ \textgreater \ 0\; \; \to \; \; (t-1)^2\ \textgreater \ 0\; \; \to \; \; t\ne 1\\\\6^{x}\ne 1\; \; \to \; \; 6^{x}\ne 6^0\; \; \to \; \; x\ne 0$

$2)\; \; 2^{x^2} \leq 64\cdot 2^{x}\\\\2^{x^2} \leq 2^6\cdot 2^{x}\\\\2^{x^2} \leq 2^{6+x}\\\\x^2 \leq 6+x\\\\x^2-x-6 \leq 0\\\\x^2-x-6=0\; \; pri\; \; x_1=-2,\; x_2=3\\\\+++[-2]---[3]+++\\\\x\in [\, -2,3\, ]$