Rút gọn
A= tanx. tan(x+pi/3)+tan(x+ pi/3).tan(x+2pi/3) +tan(x+2pi/3).tanx
Rút gọn A= tanx. tan(x+pi/3)+tan(x+ pi/3).tan(x+2pi/3) +tan(x+2pi/3).tanx
By Alexandra
By Alexandra
Rút gọn
A= tanx. tan(x+pi/3)+tan(x+ pi/3).tan(x+2pi/3) +tan(x+2pi/3).tanx
Đáp án:
Giải thích các bước giải:
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$