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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Tính giá trị biểu thức sin6a + cos6a +3sin2a . cos2a (a là alpha)

Toán Lớp 9: Tính giá trị biểu thức
sin6a + cos6a +3sin2a . cos2a (a là alpha)

Comments ( 2 )

  1. Giải đáp: 1
    Lời giải và giải thích chi tiết:
    $sin^6\alpha+cos^6\alpha+3.sin^2\alpha.cos^2\alpha$
    $=(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alpha.cos^2\alpha+cos^4\alpha)+3.sin^2\alpha.cos^2\alpha$  
    $=sin^4\alpha-sin^2\alpha.cos^2\alpha+cos^4\alpha+3.sin^2\alpha.cos^2\alpha$
    (vì $sin^2\alpha+cos^2\alpha=1$)
    $=sin^4\alpha+2sin^2\alpha.cos^2\alpha+cos^4\alpha$
    $=(sin^2\alpha+cos^2\alpha)^2$
    $=1$ (vì $sin^2\alpha+cos^2\alpha=1$)

  2. ~rai~
    \(\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha\cos^2\alpha\\=(\sin^6\alpha+\cos^6\alpha)+3\sin^2\alpha\cos^2\alpha\\=(\sin^2\alpha+\cos^2\alpha)(\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha)+3\sin^2\alpha\cos^2\alpha\\=\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha+3\sin^2\alpha\cos^2\alpha\\=(\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\sin^4\alpha)-3\sin^2\alpha\cos^2\alpha+3\sin^2\alpha\cos^2\alpha\\=(\sin^2\alpha+\cos^2\alpha)^2\\=1^2=1.\\\text{Giải thích:Công thức áp dụng:}\\+)\sin^2\alpha+\cos^2\alpha=1.\\\text{+)Hằng đẳng thức:}A^3+B^3=(A+B)(A^2-AB+B^2).\)

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222-9+11+12:2*14+14 = ? ( )