设a+b=1,a的平方加b的平方=2,则a的7次方加b的7次方=?

2个回答

  • a+b =1 两边平方得

    a^2+b^2+2ab = 1

    解得

    ab =-1/2

    (ab)^2 = 1/4

    (ab)^4 = 1/16

    a^4+b^4 = 4-2(ab)^2 = 4-1/2 = 7/2

    a^7+b^7 = (a+b)(a^7+b^7) = a^8+b^8 + ab^7+ba^7

    = (a^8+b^8)+ab(a^6+b^6)

    = (a^4+b^4)^2-2(ab)^4+ab[(a^2+b^2)(a^4+b^4-(ab)^2)]

    = (7/2)^2 - 2*(1/16)+(-1/2)*[2*(7/2 - 1/4)]

    = 71/8

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    解法2:

    a+b =1 两边平方得

    a^2+b^2+2ab = 1

    解得

    ab =-1/2

    (ab)^2 = 1/4

    (ab)^3 = -1/8

    a^3+b^3 = (a+b)(a^2-ab+b^2) = a^2-ab+b^2 = 2+1/2 = 5/2

    a^4+b^4 = (a^2+b^2)^2 - 2(ab)^2 = 4 -1/2 = 7/2

    所以

    a^7+b^7 = (a^3+b^3)(a^4+b^4)-a^3b^4-b^3a^4 = (a^3+b^3)(a^4+b^4)-(ab)^3(a+b) = (5/2)*(7/2)-(-1/8)*1 = 71/8