4^(sqrt(x-2))+16 > 10*2^(sqrt(x-2))

zarinazaac30 zarinazaac30    1   09.09.2019 22:30    0

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gilkatay1 gilkatay1  07.10.2020 04:08
4^{\sqrt{x-2}}+16\ \textgreater \ 10*2^{\sqrt{x-2}} \\ 2^{2\sqrt{x-2}}+2^4-10*2^{\sqrt{x-2}}\ \textgreater \ 0 \\
Допустим 2^{\sqrt{x-2}} = t, t\ \textgreater \ 0
Тогдаt^2-10t+16\ \textgreater \ 0 \\ \left \{ {{\left[\begin{array}{cc}t\ \textless \ 2\\t\ \textgreater \ 8\end{array}}\right \atop {t\ \textgreater \ 0}} \right. \\ 2^{\sqrt{x-2}}\ \textgreater \ 8 =\ \textgreater \ \sqrt{x-2}\ \textgreater \ 3 =\ \textgreater \ x\ \textgreater \ 11 \\ or\\ 0\ \textless \ \sqrt{x-2}\ \textless \ 1 =\ \textgreater \ 0\ \textless \ x-2\ \textless \ 1 =\ \textgreater \ 2\ \textless \ x\ \textless \ 3 \\ \\Answer: (2;3)U(11;\inf)
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