Question

найдите сумму целых решений неравенства

Answer 1

$\left(3^{2x+1}-26\cdot3^x-9\right)\cdot\log^2_3(6-x)\ \textgreater \ 0\\O.D.3:\;6-x\ \textgreater \ 0\Rightarrow x\ \textless \ 6\\\\1)\log^2_3(6-x)=0\\6-x=1\\x=5\\\log^2_3(6-x)\ \textgreater \ 0\;npu\;x\in(-\infty;\;5)\cup(5;\;+\infty)\\\\\\2)\;3^{2x+1}-26\cdot3^x-9\ \textgreater \ 0\\3\cdot3^{2x}-26\cdot3^x-9\ \textgreater \ 0\\3^x=t,\;3^2x=t^2,\;t\ \textgreater \ 0\\3t^2-26t-9\ \textgreater \ 0\\3t^2-26t-9=0\\D=676+4\cdot3\cdot9=676+108=784=(28)^2\\t_{1,2}=\frac{26\pm28}6\\t_1=9\\t_2=-\frac13\;-\;He\;nogx.\\3^x=9\\3^x=3^2\\x=2\\3\cdot3^{2x}-26\cdot3^x-9\ \textgreater \ 0\;npu\;x\ \textgreater \ 2\\KOPHu:\;3\;u\;4\\3+4=7$