решите системы неравенств:. Created by barenzi. algebra-ru

решите системы неравенств:...

barenzi @barenzi

July 2022 1 17 Report

решите системы неравенств:

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sedinalana

Verified answer

1
{5^x>25⇒x>2
{(1/3)^(x-8)<1/27⇒x-8>3⇒x>11
x∈(11;∞)
2
{8>2^(x-6)⇒x-6<3⇒x<9
3^(4x)>81⇒4x>4⇒x>1
x∈(1;9)

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1)
$\left \{ {{5^x\ \textgreater \ 25} \atop {( \frac{1}{3})^{x-8}\ \textless \ \frac{1}{27} }} \right. =\ \textgreater \ \left \{ {{5^x\ \textgreater \ 5^2} \atop {( \frac{1}{3})^{x-8}\ \textless \ ( \frac{1}{3} )^3 }} \right. =\ \textgreater \ \left \{ {{5^x\ \textgreater \ 5^2} \atop {3^{-(x-8)}\ \textless \ 3^{-3}}} \right. =\ \textgreater \ \left \{ {{x\ \textgreater \ 2} \atop {8-x\ \textless \ -3}} \right. =\ \textgreater \ \\ \\ =\ \textgreater \ \left \{ {{x\ \textgreater \ 2} \atop {x\ \textgreater \ 11}} \right. =\ \textgreater \ x \in (11; +\infty)$

2)
$\left \{ {{8\ \textgreater \ ( \frac{1}{2} )^{6-x}} \atop {3^{4x}\ \textgreater \ 81}} \right. =\ \textgreater \ \left \{ {{2^3\ \textgreater \ 2^{-(6-x)}} \atop {3^{4x}\ \textgreater \ 3^4}} \right. =\ \textgreater \ \left \{ {{3\ \textgreater \ x-6} \atop {4x\ \textgreater \ 4}} \right. =\ \textgreater \ \left \{ {{x\ \textless \ 9} \atop {x\ \textgreater \ 1}} \right. =\ \textgreater \ x\in (1;9)$

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