题目内容
计算:(1-
+
-
+…+
-
)÷(
+
+
+…+
)= .
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考点:有理数的混合运算
专题:规律型
分析:观察发现-
=
-2×
,-
=
-2×
,-
=
-2×
,…,-
=
-2×
,因而将(1-
+
-
+…+
-
)转化为[(1+
+
+
+…+
+
)-2×(
+
+
+…+
)]
再进一步转化[(1+
+
+
+…+
+
)-(1+
+
+…+
)]=(
+
+
+…+
)
至此问题解决.
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再进一步转化[(1+
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至此问题解决.
解答:解:原式=[(1+
+
+
+…+
+
)-2×(
+
+
+…+
)]÷(
+
+
+…+
),
=[(1+
+
+
+…+
+
)-(1+
+
+…+
)]÷(
+
+
+…+
),
=(
+
+
+…+
+
)÷(+
+
+
+…+
),
=1.
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=[(1+
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=(
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=1.
点评:本题目考察有理数的混合运算.解决本题的关键是巧妙利用-
=
-2×
,-
=
-2×
,-
=
-2×
,…,-
=
-2×
分数的这一特殊变化形式.
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