Решите неравенства: 3^(2x-5)+3^(2x-6)-3^(2x-7)-3^(2x-8)≤32...

$1)$

$3^{2x-5} + 3^{2x-6} - 3^{2x-7} - 3^{2x-8} \leq 32$

$3^{2x-5}(1+ 3^{-1} - 3^{-2} - 3^{-3}) \leq 32$

$3^{2x-5}(1+\frac{1}{3} } - \frac{1}{9} - \frac{1}{27} ) \leq 32$

$3^{2x-5}(1+ \frac{9}{27} } - \frac{3}{27} - \frac{1}{27} ) \leq 32$

$3^{2x-5}*1 \frac{5}{27} } \leq 32$

$3^{2x-5}*\frac{32}{27} } \leq 32$

$3^{2x-5} \leq 32* \frac{27}{32}$

$3^{2x-5} \leq 27$

$3^{2x-5} \leq 3^3$

$2x-5 \leq 3$

$2x \leq 8$

$x \leq 4$

-------------[4]-----------
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Ответ: $(-$ ∞ $;4]$

$2)$

$5^{4x} - 5^{4x-1} + 5^{4x-2} - 5^{4x-3} \geq 104$

$5^{4x} (1- 5^{-1} + 5^{-2} - 5^{-3}) \geq 104$

$5^{4x} (1- \frac{1}{5} + \frac{1}{25} - \frac{1}{125} ) \geq 104$

$5^{4x} (1- \frac{25}{125} + \frac{5}{125} - \frac{1}{125} ) \geq 104$

$5^{4x} * \frac{104}{125} \geq 104$

$5^{4x} \geq 104* \frac{125}{104}$

$5^{4x} \geq 125$

$5^{4x} \geq 5^3$

$4x \geq 3$

$x \geq 0.75$

----------[0.75]-------------
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Ответ: $[0.75;+$ ∞ $)$