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

Toán Lớp 9: Tính: `sin^6 a + cos^6 a + 3. sin^2 a . cos^2 a`

Toán Lớp 9: Tính: sin^6 a + cos^6 a + 3. sin^2 a . cos^2 a

Comments ( 2 )

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

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

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

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