Решите по братски :3

[latex]1.\;\sqrt{5-2x}=3\\O.D.3.:\\\;5x-2\geq0\\x\geq\frac25\\5-2x=9\\-2x=9-5\\-2x=4\\x=-2\\\\2.\;9^{3x+5}=\frac1{27}\\9^{3x+5}=3^{-3}\\3x+5=-3\\3x=-3-5\\3x=-8\\x=-\frac83=-2\frac23\\\\3.\;9^{\sqrt{x-5}}-27=2\cdot3^{\sqrt{x-5}+1}\\O.D.3.:\\x-5\geq0\\x\geq5\\9^{\sqrt{x-5}}-2\cdot3^{\sqrt{x-5}+1}-27=0\\3^{2\sqrt{x-5}}-6\cdot3^{\sqrt{x-5}}-27=0\\3^{\sqrt{x-5}}=t,\;3^{2\sqrt{x-5}}=t^2,\;t\ \textgreater \ 0\\t^2-6t-27=0\\D=36+4\cdot27=144=(12)^2\\t_{1,2}=\frac{6\pm12}2\\t_1=9,\\t_2=-3\;-\;He\;nogx.\\[/latex] [latex]3^{\sqrt{x-5}}=9\\3^{\sqrt{x-5}}=3^2\\\sqrt{x-5}=2\\x-5=4\\x=9\\\\4.\;\log_7\left(2x^2-2\right)=\log_7\left(-3x\right)\\O.D.3.:\\\begin{cases}2x^2-2\ \textgreater \ 0\\-3x\ \textgreater \ 0\end{cases}\Rightarrow\begin{cases}x\ \textless \ -1\\x\ \textgreater \ 1\\x\ \textless \ 0\end{cases }\Rightarrow x\ \textless \ -1\\2x^2-2=-3x\\2x^2+3x-2=0\\D=9+4\cdot2\cdot2=25\\x_{1,2}=\frac{-3\pm5}4\\x_1=-2\\x_2=\frac12\;-\;He\;nogx.[/latex]