## Rút gọn A= tanx. tan(x+pi/3)+tan(x+ pi/3).tan(x+2pi/3) +tan(x+2pi/3).tanx

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Rút gọn
A= tanx. tan(x+pi/3)+tan(x+ pi/3).tan(x+2pi/3) +tan(x+2pi/3).tanx

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1 năm 2021-10-06T14:42:22+00:00 2 Answers 3 views 0

1. Đáp án:

Giải thích các bước giải:

2. 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:

$\tan \dfrac{π}{3}.(3+A) =0$

$\Leftrightarrow A = – 3$