С-26 ВАРИАНТ 2 Решите плиз

[latex]1.\\a)\;\log_5(3x-4)=\log_5(12-5x)\\O.D.3.:\\\begin{cases}3x-4\ \textgreater \ 0\\12-5x\ \textgreater \ 0\end{cases}\Rightarrow x\in\left(1\frac13;\;2\frac25\right)\\3x-4=12-5x\\3x+5x=12-4\\8x=8\\x=1\;-\;He\;nogx.\;no\;O.D.3.\\KopHeu\;HET.\\b)\;\log_3(x^2+3x-7)=1\\x^2+3x-7=3^1\\x^2+3x-10=0\\D=9+4\cdot10=49\\x_{1,2}=\frac{-3\pm7}2\\x_1=-5,\;x_2=2[/latex] [latex]c)\;\lg(x-1)+\lg(x+1)=\lg(9x+9)\\\begin{cases}x-1\ \textgreater \ 0\\x+1\ \textgreater \ 0\\9x+9\ \textgreater \ 0\end{cases}\Rightarrow x\ \textgreater \ 1\\\lg(x-1)(x+1)=\lg(9x+9)\\x^2-1=9x+9\\x^2-9x-10=0\\D=81+4\cdot10=121\\x_{1,2}=\frac{9\pm11}2\\x_1=10,\\x_2=-1\;-\;He\;nogx.\;no\;O.D.3.\\\\2.\;\lg(3x+10)=\lg(3(4x-3)+10)\\3x+10=12x+1\\12x-3x=10-1\\9x=9\\x=1[/latex]