Помогите решить: 25^(7−5x) ≤ 0,008.. Created by crezu13. algebra-ru

Помогите решить: 25^(7−5x) ≤ 0,008...

August 2022 2 4 Report

Помогите решить: 25^(7−5x) ≤ 0,008.

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$25^{7-5x} \leq 0,008 \\ \\ (5^2)^{7-5x} \leq \frac{1}{125} \\ \\ 5^{14-10x} \leq \frac{1}{5^3} \\ \\\ 5^{14-10x} \leq 5^{-3} \\ \\ 14-10x \leq -3 \\ -10x \leq -3-14 \\ -10x \leq -17 \\ 10x \geq 17 \\ x \geq 1.7 \\ \\ x \in [1.7; +\infty)$

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fetibuy 175-125x 0.008
-125x 0.008-175
-125x -174.002
x -174.002+125
x -49.002

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