Решить неравенство [tex] 4^{x} -6*2 ^{x} +8 \geq 0[/tex]. Created by vlusunova. algebra-ru

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nelle987

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$4^x-6\cdot2^x+8\geqslant0\\
2^x=t>0:\quad t^2-6t+8\geqslant0\\
2^x=t>0:\quad (t-2)(t-4)\geqslant0\\
t\in(0,2]\cup[4,+\infty)\\
2^x\in(0,2^1]\cup[2^2,+\infty)\\
x\in(-\infty,1]\cup[2,+\infty)$

1 votes Thanks 1

darkhan91kz $2^{2x} ( 2^{2} -6) \geq 2^{3}$
дальше ерунда переходи к показателям и решай

0 votes Thanks 1

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