用数学归纳法证明当n是正奇数时,xn+yn能被x+y整除. 答案:证明:(1)当n=1时,xn+yn=x+y能被x+y整除. (2)假设n=2k-1时命题成立,即(x2k-1+y2k-1)能被x+y整除,那么n=2k+1时有x2k+1+y2k+1=x2k-1·x2+y2k-1·y2 =x2k-1·x2+x2k-1·y2-x2k-1·y2+y2k-1·y2 =x2k-1(x2-y2)+(x2k-1+y2k-1)·y2 =x2k-1(x+y)(x-y)+(x2k-1+y2k-1)·y2. ∵x2k-1(x+y)(x-y)与(x2k-1+y2k-1)y2都能被x+y整除,∴n=按要求转换句型
Does her father like carrots?(作否定回答)
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