题目内容
用简便方法计算:
(1+
+
+
)×(
+
+
+
)-(1+
+
+
+
)×(
+
+
)=
.
(1+
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
5 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
5 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
6 |
1 |
6 |
分析:通过仔细观察,此题数字具有一定特点,可用字母代替数的方法解答,可设
+
+
=a,
+
+
+
=b,把a、b代入原式,解答即可.
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
5 |
解答:解:设
+
+
=a,
+
+
+
=b,则:
(1+
+
+
)×(
+
+
+
)-(1+
+
+
+
)×(
+
+
),
=(1+a)×b-(1+b)×a,
=b+ab-a-ab,
=b-a,
=(
+
+
+
)-(
+
+
),
=
.
故答案为:
.
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
6 |
(1+
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
5 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
5 |
1 |
2 |
1 |
3 |
1 |
4 |
=(1+a)×b-(1+b)×a,
=b+ab-a-ab,
=b-a,
=(
1 |
2 |
1 |
3 |
1 |
4 |
1 |
6 |
1 |
2 |
1 |
3 |
1 |
4 |
=
1 |
6 |
故答案为:
1 |
6 |
点评:凡是这类问题,可采用用字母代替数的方法,使运算简便.
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