Question

даю 50 баллов, надо решить неравенство ​

Answer 1

$\displaystyle 1) (\sqrt[3]{3})^x^+^6>\frac{1}{9}\\\\3^{\frac{1}{3}(x+6) }>3^-^2\\\\3^{\frac{1}{3}x+2 }>3^-^2\\\\\frac{1}{3}x+2>-2\\\\x+6>-6\\x>-6-6\\x>-12, x \in (-12, + \infty)\\\\2)(1\frac{2}{7})^{x^2-4}\leq 1\\\\(\frac{9}{7})^{x^2-4}\leq 1\\\\(\frac{9}{7})^{x^2-4}\leq (\frac{9}{7})^0\\\\x^2-4\leq 0\\x^2\leq 4\\|x|\leq 2\\x\leq 2,x\geq 0 = x \in [0,2]\\-x\leq 2,x<0=x\geq -2,x<0=x \in [-2,0)\\x\in [-2,2]$

Answer 2

Ответ:

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