Đáp án:

 $A=-3$

Giải thích các bước giải:

Áp dụng công thức:

$ \tan (a - b) = \dfrac{\tan a - \tan b}{1 + \tan a.\tan b}$

$ ⇒ \tan (a - b).(1 + \tan a.\tan b) = \tan a - \tan b$ (*)

$A=\tan x.\tan \left({x + \dfrac{π}{3}}\right) + \tan \left({x + \dfrac{π}{3}}\right).\tan \left({x + \dfrac{2π}{3}}\right) + \tan \left({x + \dfrac{2π}{3}}\right).\tan x$

Áp dụng (*) ta có:

+)

$ \tan \dfrac{π}{3}.\left[{1 + \tan x.\tan \left({x + \dfrac{π}{3}}\right)}\right]$

$ = \tan \left[{\left({x + \dfrac{π}{3}}\right) - x)}\right].\left[{1 + \tan x.\tan \left({x + \dfrac{π}{3}}\right)}\right]$

$= \tan \left({x + \dfrac{π}{3}}\right) - \tan x$ (1)

+)

$\tan \dfrac{π}{3}.\left[{1 + \tan \left({x + \dfrac{π}{3}}\right).\tan \left({x + \dfrac{2π}{3}}\right)}\right]$

$= \tan \left[{\left({x + \dfrac{2π}{3}}\right) - \left({x + \dfrac{π}{3}}\right)}\right].\left[{1 + \tan \left({x + \dfrac{π}{3}}\right).\tan \left({x + \dfrac{2π}{3}}\right)}\right] $

$ = \tan \left({x + \dfrac{2π}{3}}\right) - \tan \left({x + \dfrac{π}{3}}\right)$ (2)

+)

$ \tan \dfrac{π}{3}.\left[{1 + \tan \left({x + \dfrac{2π}{3}}\right).\tan x}\right] $

$ = \tan \dfrac{- 2π}{3}.\left[{1 + \tan \left({x + \dfrac{2π}{3}}\right).\tan x}\right] $

$ = \tan \left[{x - \left({x + \dfrac{2π}{3}}\right)}\right].\left[{1 + \tan \left({x + \dfrac{π}{3}}\right).\tan x}\right]$

$ = \tan x - \tan \left({x + \dfrac{2π}{3}}\right)$ (3)

Lấy vế cộng vế phương trình (1) + (2) + (3) ta có:

$ \tan \dfrac{π}{3}.(3+A) =0$

$ \Leftrightarrow A = - 3$