помогите решить математику- в приложении

$1) \ log_{1/3}(3x - 6) = -2\\\\log_{1/3}(3x - 6) = log_{1/3}(9)\\\\3x-6 = 9\\\\x=5\\\\ 1') \ 3^{2-x} = 81^x\\\\ 3^{2-x} = 3^{4x}\\\\ 2-x = 4x\\\\x=0.4\\\\ 2) f'(1) - ?\\\\f(x)= (3-2x)/(x+1)\\\\f'(x) = ((3-2x)/(x+1))' =\\ ((x+1)(3-2x)' - (3-2x)(x+1)')/(x+1)^2 =\\(-2x-2-3+2x)/(x+1)^2 = -5/(x+1)^2\\\\f'(1) = -5/4$

2}(2-x)/(4-x^2) - ?, \ lim_{x - > 2}(2-x)-> 0, \\ lim_{x - > 2}(4-x^2)-> 0\\\\ f(x) = (2-x), f'(x) = -1, g(x) = (4-x^2), g'(x) = -2x" alt="3) lim_{x - > 2}(2-x)/(4-x^2) - ?, \ lim_{x - > 2}(2-x)-> 0, \\ lim_{x - > 2}(4-x^2)-> 0\\\\ f(x) = (2-x), f'(x) = -1, g(x) = (4-x^2), g'(x) = -2x" align="absmiddle" class="latex-formula">

lim_{x -> 2}(f'(x)/g'(x)) = lim_{x -> 2}1/2x = 1/4

Использовали правило Лопиталя.