Пожалуйста,помогите. Алгебра

1) [latex] \left \{ {{3(x-1)-2(1+x)\ \textless \ 1} \atop {3x-4\ \textgreater \ 0}} \right. [/latex] [latex] \left \{ {{3x-3-2-2x\ \textless \ 1} \atop {3x\ \textgreater \ 4}} \right. [/latex] [latex] \left \{ {{x\ \textless \ 6} \atop {x\ \textgreater \ \frac{4}{3} }} \right. [/latex] x∈(4/3;6) 2) (√6+√3)√12-2√6*√3=√72+√36-2√18=6√2+6-6√2=6 3)а) [latex] 6^{-2} + ( \frac{9}{4} ) ^{-1} = \frac{1}{6 ^{2} } + \frac{1}{( \frac{9}{4}) } = \frac{1}{36} + \frac{4}{9} = \frac{1}{36} + \frac{16}{36} = \frac{15}{36} = \frac{5}{12} [/latex] б)[latex](2 ^{-3} ) ^{2} = 2^{-6} = \frac{1}{2 ^{6} } = \frac{1}{y} = \frac{1}{64} [/latex] в)[latex] \frac{7 ^{-8} * 7 ^{-9} }{7 ^{-15} } = \frac{7 ^{-17} }{ 7 ^{-15} } =7 ^{-2} = \frac{1}{7 ^{2} } = \frac{1}{49} [/latex] 4)[latex]( \frac{3}{9 - x^{2} } + \frac{1}{x-3} ) : \frac{x}{ x^{2} - 6x +9} =( \frac{3}{(3-x)(3+x)} - \frac{3+x}{(3-x)(3+x)} ) : \frac{x}{(x-3) ^{2} } = [/latex][latex] \frac{x}{(3+x)(3-x)} * \frac{(3-x) ^{2} }{x} = \frac{3-x}{3+x} [/latex] 6)[latex]- \frac{x-8}{4} +1 \geq 0;   \frac{x-8}{4} \leq 1;  x-8 \leq 4; x \leq 12[/latex] x∈(-∞;12]