Question

помогите решить математику- в приложении

Answer 1

 

$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$

 

$3) lim_{x - > 2}(2-x)/(4-x^2) - ?, lim_{x - > 2}(2-x)-> 0, \ lim_{x - > 2}(4-x^2)-> 0\\ f(x) = <var></var> (2-x), f'(x) = -1, g(x) = (4-x^2), g'(x) = -2x<var><var>$

 

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

 

Использовали правило Лопиталя.