(1+
1
2
)×(1-
1
2
)×(1+
1
3
)×(1-
1
3
)×…×(1+
1
100
)×(1-
1
100
分析:首先把括號內(nèi)的加減運算完成,然后前后分子分母約分,即可得解.
解答:解:(1+
1
2
)×(1-
1
2
)×(1+
1
3
)×(1-
1
3
)×…×(1+
1
100
)×(1-
1
100

=
3
2
×
1
2
×
4
3
×
2
3
×
5
4
×
3
4
×…×
101
100
×
99
100

=
1
2
×
3
2
×
2
3
×
4
3
×
3
4
×
5
4
×…×
99
100
×
101
100

=
1
2
×
101
100

=
101
200
點評:利用分?jǐn)?shù)的分?jǐn)?shù)的四則混合運算,首先計算括號內(nèi)的,然后約分是解決此題的關(guān)鍵.
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