Помогите пожалуйста)

0\\ x>3\\ (3;+oo)\\ " alt="y=log_{3}(x(x-3))-log_{3}x\\ y=log_{3}(\frac{x(x-3)}{x})\\ y=log_{3}(x-3)\\ x-3>0\\ x>3\\ (3;+oo)\\ " align="absmiddle" class="latex-formula">

130\\ 5^{\frac{1}{x}}+5^{\frac{1}{x}}*25>130\\ 26*5^{\frac{1}{x}}>130\\ 5^{\frac{1}{x}}>5\\ \frac{1}{x}>1\\ x>1\\ " alt="5^{\frac{1}{x}}+5^{\frac{1}{x}+2}>130\\ 5^{\frac{1}{x}}+5^{\frac{1}{x}}*25>130\\ 26*5^{\frac{1}{x}}>130\\ 5^{\frac{1}{x}}>5\\ \frac{1}{x}>1\\ x>1\\ " align="absmiddle" class="latex-formula">

$9^{-x} -28*3^{-(x+1)}+3<0\\ \frac{1}{3^{2x}}-28*\frac{1}{3*3^x}+3<0\\ \frac{1}{3^{2x}}=a\\ a^2-\frac{28}{3}*a+3<0\\ D=\frac{28^2}{3^2}-4*1*3=\frac{26^2}{3^2}\\ a=\frac{1}{3}\\ a=9\\ 3^{-2x}=3^2\\ x=-1\\ 3^{-2x}=3^{-1}\\ x=1\\ (-oo;1)$

$3^{-x-3} \geq 27\\ 3^{-x-3} \geq 3^3\\ -x-3 \geq 3\\ -x \geq 6\\ x \leq -6\\ (-oo;-6]$

0\\ x>-1\\ 7-x^2>0\\ -x^2>-7\\ x^2<7\\ (-1;\sqrt{7})\\ " alt="y=\frac{ln(7-x^2)}{x+1}\\ \\ x+1>0\\ x>-1\\ 7-x^2>0\\ -x^2>-7\\ x^2<7\\ (-1;\sqrt{7})\\ " align="absmiddle" class="latex-formula"> сумма 1+2=3

0\\ D=36+4*1*9 = 6\sqrt{2}\\ x_{1}=\frac{1+\sqrt{2}}{3}\\ x_{2}=\frac{1-\sqrt{2}}{3}\\ (-oo;\frac{1-\sqrt{2}}{3}) U (\frac{1+\sqrt{2}}{3};+oo)" alt="9x^2-6x-1>0\\ D=36+4*1*9 = 6\sqrt{2}\\ x_{1}=\frac{1+\sqrt{2}}{3}\\ x_{2}=\frac{1-\sqrt{2}}{3}\\ (-oo;\frac{1-\sqrt{2}}{3}) U (\frac{1+\sqrt{2}}{3};+oo)" align="absmiddle" class="latex-formula">

1\\ 3-6x>0\\ -6x>-3\\ x<\frac{1}{2}\\ " alt="2^{3-6x}>1\\ 3-6x>0\\ -6x>-3\\ x<\frac{1}{2}\\ " align="absmiddle" class="latex-formula">
очевидно что это -1