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已知x2=x+1,y2=y+1,且x≠y.
(1)求证:x+y=1;(2)求x5+y5的值.
人气:473 ℃ 时间:2020-02-06 02:03:37
解答
(1)证明:∵x2=x+1,y2=y+1,
∴x2-y2=x-y
∴x+y=1(x≠y)
(2)∵x2=x+1,y2=y+1,∴x3=x2+x,y3=y2+y,x4=x3+x2,y4=y3+y2,x5=x4+x3,y5=y4+y3
∴x5+y5
=x4+x3+y4+y3
=x3+x2+x2+x+y3+y2+y2+y,
=x2+x+x2+x2+x+y2+y+y2+y2+y,
=3(x2+y2)+2(x+y),
=3(x+1+y+1)+2(x+y),
=3×3+2,
=11.
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