Question

срочно нужна помощь, решите неравенство $sqrt[3]{5} geq 25^{x+2}$

Answer 1

$sqrt[3]{5} geq 25^{(x+2)}$
$5^{ frac{1}{3}} geq (5^{2})^{(x+2)}$
$5^{ frac{1}{3}} geq 5^{(2x+4)}$
$frac{1}{3}} geq 2x+4$
$1 geq 6x+12$
$x leq - frac{11}{6}$

Ответ: $x leq - 1frac{5}{6}$

Answer 2

$sqrt[3]{5}geq25^{x+2}\5^{frac{1}{3}}geq5^{2(x+2)}\frac{1}{3}geq2x+4\-frac{11}{3}geq2x\-frac{11}{6}geq x$

\\\\\\\\\\\\\\α\\\\β////////////
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где $alpha=-frac{11}{6}$, $beta$ – 0. 
x∈(–∞; $-frac{11}{6}$]

Answer 3

а откуда -11/3?

Answer 4

1/3 – 4 = –3 2/3 = –11/3