Câu hỏi 27

\(\eqalign{  & f\left( x \right) = \tan \left( {x - {{2\pi } \over 3}} \right) = {{\tan x - \tan {{2\pi } \over 3}} \over {1 + \tan x.\tan {{2\pi } \over 3}}} = {{\tan x + \sqrt 3 } \over {1 - \sqrt 3 \tan x}}  \cr   & f'\left( x \right) = {{\left( {\tan x + \sqrt 3 } \right)'\left( {1 - \sqrt 3 \tan x} \right) - \left( {\tan x + \sqrt 3 } \right)\left( {1 - \sqrt 3 \tan x} \right)'} \over {{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}  \cr   & f'\left( x \right) = {{{1 \over {{{\cos }^2}x}}\left( {1 - \sqrt 3 \tan x} \right) - \left( {\tan x + \sqrt 3 } \right)\left( { - {{\sqrt 3 } \over {{{\cos }^2}x}}} \right)} \over {{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}  \cr   & f'\left( x \right) = {{{1 \over {{{\cos }^2}x}} = {{\sqrt 3 \tan x} \over {{{\cos }^2}x}} + {{\sqrt 3 \tan x} \over {{{\cos }^2}x}} + {3 \over {{{\cos }^2}x}}} \over {{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}  \cr   & f'\left( x \right) = {4 \over {{{\cos }^2}x{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}  \cr   &  \Rightarrow f'\left( 0 \right) = {4 \over {1\left( {1 - \sqrt 3 .0} \right)}} = 4 \cr} \)

Chọn B.