题目内容
(2013?东城区模拟)计算下面各题,能简便计算的要简便计算
(1)7
-(2
-2.3)
(2)4.85×3
-3.6+6.15×3
(3)0.025×999×2.8×40÷2
(4)
(5)(1-
)(1-
)(1-
)…(1-
).
(1)7
4 |
15 |
4 |
15 |
(2)4.85×3
3 |
5 |
3 |
5 |
(3)0.025×999×2.8×40÷2
4 |
5 |
(4)
2001×2001+2001 |
2002×2002-2002 |
(5)(1-
1 |
2×2 |
1 |
3×3 |
1 |
4×4 |
1 |
2001×2001 |
分析:(1)先去掉括号,再计算;
(2)运用乘法分配律简算;
(3)运用乘法结合律简算;
(4)把分子和分母都运用乘法分配律简算,然后再化简;
(5)1-
=1-
=
=
×
;
1-
=1-
=
=
×
;
…
由此化简求解.
(2)运用乘法分配律简算;
(3)运用乘法结合律简算;
(4)把分子和分母都运用乘法分配律简算,然后再化简;
(5)1-
1 |
2×2 |
1 |
4 |
3 |
4 |
1 |
2 |
3 |
2 |
1-
1 |
3×3 |
1 |
9 |
8 |
9 |
2 |
3 |
4 |
3 |
…
由此化简求解.
解答:解:(1)7
-(2
-2.3),
=7
-2
+2.3,
=5+2.3,
=7.3;
(2)4.85×3
-3.6+6.15×3
,
=4.85×3.6-3.6+6.15×3.6,
=(4.85-1+6.15)×3.6,
=10×3.6,
=36;
(3)0.025×999×2.8×40÷2
,
=(0.025×40)×(2.8÷2.8)×999,
=1×1×999,
=999;
(4)
,
=
,
=
,
=1;
(5)(1-
)(1-
)(1-
)…(1-
),
=(
×
)×(
×
)×(
×
)×…(
×
),
═
×(
×
)×(
×
)×…(
×
)×
,
=
×1×1×…×
,
=
.
4 |
15 |
4 |
15 |
=7
4 |
15 |
4 |
15 |
=5+2.3,
=7.3;
(2)4.85×3
3 |
5 |
3 |
5 |
=4.85×3.6-3.6+6.15×3.6,
=(4.85-1+6.15)×3.6,
=10×3.6,
=36;
(3)0.025×999×2.8×40÷2
4 |
5 |
=(0.025×40)×(2.8÷2.8)×999,
=1×1×999,
=999;
(4)
2001×2001+2001 |
2002×2002-2002 |
=
2001×(2001+1) |
2002×(2002-1) |
=
2001×2002 |
2002×2001 |
=1;
(5)(1-
1 |
2×2 |
1 |
3×3 |
1 |
4×4 |
1 |
2001×2001 |
=(
1 |
2 |
3 |
2 |
2 |
3 |
4 |
3 |
3 |
4 |
5 |
4 |
2000 |
2001 |
2002 |
2001 |
═
1 |
2 |
3 |
2 |
2 |
3 |
4 |
3 |
3 |
4 |
2001 |
2000 |
2000 |
2001 |
2002 |
2001 |
=
1 |
2 |
2002 |
2001 |
=
1001 |
2001 |
点评:此题考查四则混合运算,要仔细观察算式的特点,灵活运用一些定律进行简便计算;较复杂的计算要先找出规律,然后根据规律化简求解.
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