[Toán 11]

H

hn3

[TEX]cosx(2+5cos5x+cosx)=4[/TEX]

[TEX]<=> 2cosx+5cosxcos5x+cos^2x-4=0[/TEX]

[TEX]<=> 4cosx+5cos6x+5cos4x+2cos^2x-8=0[/TEX]

Xong sao nữa nhẩy :-S Làm nhầm hay cách khác chăng 8-|
 
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N

niemkieuloveahbu

Bài này định đánh giá,khổ nỗi là nó trái dấu nhau,>"<.

Giả như [TEX]VT=(cosx+1)^2+5(cosxcos5x-1)[/TEX]
 
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G

genius_hocmai

cosx (2+5cos5x +cosx)=4
[tex] \Leftrightarrow cos^2x +2cosx+ 5cos5x.cosx =4 [/tex]
[tex]\Leftrightarrow (cosx+1)^2 +5(cos5x.cosx -1)=0[/tex]
[tex]\Leftrightarrow (cosx+1)^2 +5(\frac{cos6x+cos2x}{2}-1)=0[/tex]
[tex]\Leftrightarrow(cosx+1)^2 +5(\frac{4{cos2x}^3+2{cosx}^2 -1}{2} -1)[/tex]=0
[tex] (cosx+1)^2+ \frac{5}{2}(4{cos2x}^3 +2{cos2x}^2-3cos2x -3 )=0[/tex]
[tex]2(cosx+1)^2 +5(cos2x-1)[/tex].([tex]4{cos2x}^2 +6cos2x -3)=0[/tex]
[tex] 2(cosx+1)^2 +10({cosx}^2-1}[/tex][tex](4{cos2x}^2 +6cos2x -3)=0[/tex]
[tex](cosx+1).(2cosx+2+........)[/tex]
 
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