1.If the equation m(x-1)=2001-n(x-2) for x has infinite root

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1.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=( ).
（equation方程 infinite roots无数个根）
2.We have the following numbers 9/5 12/7 27/17 36/19 54/29,the maximum number among them is( ),the minimum number is( ).
(maximum 最大的 minimum最小的）
3.The sequence 1/1 1/2 2/2 1/3 2/3 3/3 1/4 2/4 3/4 4/4 1/5 2/5.,then the 2011th number is( )
(sequence 序列)
题目都看懂了,.
能做一道是一道,有过程最好,直接答案也行.

1.化简该式为 (m+n)x=2001+2n+m
因为该式中x有无数个解,则(m+n)=0,且2001+2n+m=0
所以,m=-n,m=2001,n=-2001
所以 m的2001次方+n的2001次方=0
2.化简即可,知the maximum is 36/19,the minimum number is 27/17
3.依据等差数列求和,可知,the 2011th number is 57/63
上面那位错了,计算错误 ,第2题,没有比较12/7和27/17的大小,实际上,27/17最小；
第3题,如果反代45进去的话,根本到不了第2011个数值,只能到1035个数值.

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