[toán11]các bạn ơi giải dùm minh cai nghĩ hok ra

D

doremon.

[tex]cos^4x+ 2sin^4x - cos^2(2x) - 3sin4x = -2[/tex]
\Leftrightarrow[TEX]cos^4x+2sin^4x-(cos^2x-sin^2x)^2-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]sin^4x+2sin^2xcos^2x-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]sin^4x+2(1-sin^2x)sin^2x-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]2sin^2x-sin^4x-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]sin^2x(2-sin^2x)-3sin4x=-2[/TEX]
\Leftrightarrow[TEX](1-cos2x)(3+cos2x)-12sin4x=-8[/TEX]
\Leftrightarrow[TEX]11-2cos2x-cos^22x-24sin2xcos2x=0[/TEX]
đặt t=tanx,[TEX]cosx \neq 0[/TEX]
\Rightarrow[TEX]\frac{(1-t^2)^2}{(1+t^2)^2}+\frac{2-2t^2}{1+t^2}+\frac{48t(t-t^2)}{1+t^2)^2}-11=0[/TEX]
\Leftrightarrow[TEX]12t^4+48t^3+24t^2-48t+8=0[/TEX]
 
N

ngomaithuy93

dây nè: [tex]cos^4x+ 2sin^4x - cos^2(2x) - 3sin4x = -2[/tex]
Bạn chú ý đánh tex+tiêu đề ná;)
[tex]cos^4x+ 2sin^4x - cos^22x - 3sin4x = -2[/tex]
\Leftrightarrow[TEX]\frac{(1+cos2x)^2}{4}+\frac{(1-cos2x)^2}{4}-cos^22x-6sin2xcos2x+2=0[/TEX]
\Leftrightarrow[TEX] -2cos^22x-6sin2xcos2x+4=0[/TEX]
\Leftrightarrow[TEX]2cos^22x-6sin2xcos2x+4sin^22x=0[/TEX]
 
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H

ha_trung2003

\Leftrightarrow[TEX]cos^4x+2sin^4x-(cos^2x-sin^2x)^2-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]sin^4x+2sin^2xcos^2x-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]sin^4x+2(1-sin^2x)sin^2x-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]2sin^2x-sin^4x-3sin4x=-2[/TEX]
\Leftrightarrow[TEX]sin^2x(2-sin^2x)-3sin4x=-2[/TEX]
\Leftrightarrow[TEX](1-cos2x)(3+cos2x)-12sin4x=-8[/TEX]
\Leftrightarrow[TEX]11-2cos2x-cos^22x-24sin2xcos2x=0[/TEX]
đặt t=tanx,[TEX]cosx \neq 0[/TEX]
\Rightarrow[TEX]\frac{(1-t^2)^2}{(1+t^2)^2}+\frac{2-2t^2}{1+t^2}+\frac{48t(t-t^2)}{1+t^2)^2}-11=0[/TEX]
\Leftrightarrow[TEX]12t^4+48t^3+24t^2-48t+8=0[/TEX]
bạn làm kiểu j mình chẳng hỉu j cả
hình như bạn làm sai rùi hay sao ấy
 
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