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[nxt96]

GIẢI PHƯƠNG TRÌNH

 

$\large tan^{2}2x*tan^{2}3x*tan5x=tan^{2}2x-tan^{2}3x+tan5x$

Bài viết đã được chỉnh sửa nội dung bởi nxt96: 16-06-2013 - 09:17

[sieumau88]

Ta có___ $ tan5x = \dfrac{tan2x + tan3x}{1-tan2x.tan3x}$

Do đó__ $ tan5x \ . \ \left ( 1 - tan2x.tan3x \right ) = tan2x + tan3x$
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$\boxed{tan^{2}2x \ . \ tan^{2}3x \ . \ tan5x = tan^{2}2x-tan^{2}3x+tan5x}$

ĐKXĐ : $\left\{\begin{matrix}
tan2x.tan3x \neq 1 \\
cos2x \neq 0 \\
cos3x \neq 0 \\
cos5x \neq 0
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
cosx \neq 0 \\
cos2x \neq 0 \\
cos3x \neq 0 \\
cos5x \neq 0
\end{matrix}\right.$

pt $\Leftrightarrow tan^{2}3x-tan^{2}2x = tan5x \ . \ \left ( 1 - tan^{2}2x \ . \ tan^{2}3x \right )$

$\Leftrightarrow tan^{2}3x-tan^{2}2x = tan5x \ . \ \left ( 1 - tan2x.tan3x \right ) \ . \ \left ( 1+tan2x.tan3x \right )$

$\Leftrightarrow \left (tan3x+tan2x \right ) \ . \ \left (tan3x-tan2x \right ) = \left (tan2x+tan3x \right ) \ . \ \left ( 1+tan2x.tan3x \right )$
 
$\Leftrightarrow \left[ \begin{array}{l} tan2x+tan3x=0\\ tan3x-tan2x = 1+tan2x.tan3x \\ \end{array} \right.$
 
$\Leftrightarrow \left[ \begin{array}{l} tan3x= - tan2x \\ \dfrac{tan3x-tan2x}{1+tan2x.tan3x} = 1 \\ \end{array} \right.$

(Vi` :__ $1+tan2x.tan3x = \dfrac{cos2x.cos3x+sin2x.sin3x}{cos2x.cos3x}= \dfrac{cos5x}{cos2x.cos3x} \neq 0 $ )

 
$\Leftrightarrow \left[ \begin{array}{l} tan3x=tan \left(-2x \right) \\ tanx = tan \dfrac{\pi}{4} \\ \end{array} \right.$
 
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