Question

Пожалуйста,помогите.
Алгебра

Answer 1

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