If the equation m(x-1)=2001-n(x-2) for x has infinite roots
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If the equation m(x-1)=2001-n(x-2) for x has infinite roots ,then m的2001次方+n的2001次方=
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展开得mx-m=2001-nx+2n
整理得(m+n)x=2001+2n+m
因为x有infinite roots,即无限解,所以方程与x无关,即m+n=0
所以有m+n=0且2001+2n+m=0
两式联立,得m=2001,n=-2001
所以m^2001+n^2001=2001^2001+(-2001)^2001=0
结果是零咯
整理得(m+n)x=2001+2n+m
因为x有infinite roots,即无限解,所以方程与x无关,即m+n=0
所以有m+n=0且2001+2n+m=0
两式联立,得m=2001,n=-2001
所以m^2001+n^2001=2001^2001+(-2001)^2001=0
结果是零咯
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