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1988
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Problem 15
AMC 8 1988 Problem 15
AMC8 1988 年第 15 题
表达式与运算
★★★☆☆
The reciprocal of
(
1
2
+
1
3
)
\displaystyle \left( \frac{1}{2}+\frac{1}{3}\right)
(
2
1
+
3
1
)
is
(
1
2
+
1
3
)
\displaystyle \left( \frac{1}{2}+\frac{1}{3}\right)
(
2
1
+
3
1
)
的倒数是
A.
1
6
\displaystyle \frac{1}{6}
6
1
B.
2
5
\displaystyle \frac{2}{5}
5
2
C.
6
5
\displaystyle \frac{6}{5}
5
6
D.
5
2
\displaystyle \frac{5}{2}
2
5
E.
5
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
1
2
+
1
3
=
5
6
\displaystyle \frac{1}{2} + \frac{1}{3} = \frac{5}{6}
2
1
+
3
1
=
6
5
, and the reciprocal of
5
6
\displaystyle \frac{5}{6}
6
5
is
6
5
\displaystyle \frac{6}{5}
5
6
.
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