题目内容
简便计算.
(1)
+
+
+…+
+
(2)1994
×79+
×790+244.9
(3)1÷(2÷3)÷(3÷4)÷(4÷5)÷…÷(2002÷2003)÷(2003÷2004)
(1)
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
48×49 |
1 |
49×50 |
(2)1994
1 |
2 |
6 |
25 |
(3)1÷(2÷3)÷(3÷4)÷(4÷5)÷…÷(2002÷2003)÷(2003÷2004)
分析:(1)每个分数的分母都是两个连续自然数的乘积,于是把每个分数拆成两个分数相减的形式,然后通过加减相互抵消,求出结果;
(2)把分数化为小数,把244.9看作79×3.1,把原式变为1994.5×79+2.4×79+79×3.1,运用乘法分配律简算;
(3)把括号内的除法算式改为分数形式,然后再把除法改为乘法,约分即可.
(2)把分数化为小数,把244.9看作79×3.1,把原式变为1994.5×79+2.4×79+79×3.1,运用乘法分配律简算;
(3)把括号内的除法算式改为分数形式,然后再把除法改为乘法,约分即可.
解答:解:(1)
+
+
+…+
+
,
=1-
+
-
+
-
+…+
-
+
-
,
=1-
,
=
;
(2)1994
×79+
×790+244.9,
=1994.5×79+0.24×790+79×3.1,
=1994.5×79+2.4×79+79×3.1,
=(1994.5+2.4+3.1)×79,
=2000×79,
=158000;
(3)1÷(2÷3)÷(3÷4)÷(4÷5)÷…÷(2002÷2003)÷(2003÷2004),
=1÷
÷
÷
÷…÷
÷
,
=1×
×
×
×…×
×
,
=1002.
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
48×49 |
1 |
49×50 |
=1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
48 |
1 |
49 |
1 |
49 |
1 |
50 |
=1-
1 |
50 |
=
49 |
50 |
(2)1994
1 |
2 |
6 |
25 |
=1994.5×79+0.24×790+79×3.1,
=1994.5×79+2.4×79+79×3.1,
=(1994.5+2.4+3.1)×79,
=2000×79,
=158000;
(3)1÷(2÷3)÷(3÷4)÷(4÷5)÷…÷(2002÷2003)÷(2003÷2004),
=1÷
2 |
3 |
3 |
4 |
4 |
5 |
2002 |
2003 |
2003 |
2004 |
=1×
3 |
2 |
4 |
3 |
5 |
4 |
2003 |
2002 |
2004 |
2003 |
=1002.
点评:注意分析式中数据,运用运算定律或运算技巧,灵活解答.
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