题目内容
(1)(987+4097×986)÷(987×4097-3110)
(2)
+
+
+…+
(3)
+
+
+…+
(4)2011×20122012-2012×20112011
(5)(5
-0.8+2
)×(7.6÷
+2
×1.25)
(6)
(7)(1+
+
+
)×(
+
+
+
)-(1+
+
+
+
)×(
+
+
)
(2)
1 |
2 |
1 |
4 |
1 |
8 |
1 |
256 |
(3)
1 |
1×4 |
1 |
4×7 |
1 |
7×10 |
1 |
97×100 |
(4)2011×20122012-2012×20112011
(5)(5
5 |
9 |
4 |
9 |
4 |
5 |
2 |
5 |
(6)
1×3×24+2×6×48+3×9×72 |
1×2×4+2×4×8+3×6×12 |
(7)(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 |
分析:(1)把第二个括号中的987看作986+1,运用乘法分配律简算,前后两个括号算式相同,因此结果为1;
(2)通过观察,可把每个分数都可以拆成两个分数相减的形式,然后通过加减相互抵消,求得结果;
(3)每个分数的分母中的两个因数相差3,于是原式变为
×(1-
+
-
+
-
+…+
-
),然后通过加减相互抵消,求得结果;
(4)把20122012拆成2012×10001,把2012×20112011拆成2011×10001,发现减号两边算式相同,股结果为0;
(5)两个括号同时计算,第一个括号运用加法交换律与结合律简算,第二个括号运用乘法分配律简算;
(6)分数的分子与分母运用乘法分配律简算,然后约分即可;
(7)此题可设1+
+
+
=a,
+
+
=b,把字母代入原式,使计算简便.
(2)通过观察,可把每个分数都可以拆成两个分数相减的形式,然后通过加减相互抵消,求得结果;
(3)每个分数的分母中的两个因数相差3,于是原式变为
1 |
3 |
1 |
4 |
1 |
4 |
1 |
7 |
1 |
7 |
1 |
10 |
1 |
97 |
1 |
100 |
(4)把20122012拆成2012×10001,把2012×20112011拆成2011×10001,发现减号两边算式相同,股结果为0;
(5)两个括号同时计算,第一个括号运用加法交换律与结合律简算,第二个括号运用乘法分配律简算;
(6)分数的分子与分母运用乘法分配律简算,然后约分即可;
(7)此题可设1+
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
解答:解:(1)(987+4097×986)÷(987×4097-3110),
=(987+4097×986)÷(986×4097+4097-3110),
=(987+4097×986)÷(4097×986+987),
=1;
(2)
+
+
+…+
,
=1-
+
-
+
-
+…+
-
,
=1-
,
=
;
(3)
+
+
+…+
,
=
×(1-
+
-
+
-
+…+
-
),
=
×(1-
),
=
×
,
=
;
(4)2011×20122012-2012×20112011,
=2011×2012×10001-2012×2011×10001,
=0;
(5)(5
-0.8+2
)×(7.6÷
+2
×1.25),
=(8-0.8)×(7.6×1.25+2.4×1.25),
=7.2×[(7.6+2.4)×1.25],
=7.2×12.5,
=0.9×(8×12.5),
=0.9×100,
=90;
(6)
,
=
,
=9;
(7)设1+
+
+
=a,
+
+
=b,则:a-b=1+
+
+
-(
+
+
)=1,
(1+
+
+
)×(
+
+
+
)-(1+
+
+
+
)×(
+
+
),
=a×(b+
)-(a+
)×b,
=ab+
a-ab-
b,
=
×(a-b),
=
×1,
=
.
=(987+4097×986)÷(986×4097+4097-3110),
=(987+4097×986)÷(4097×986+987),
=1;
(2)
1 |
2 |
1 |
4 |
1 |
8 |
1 |
256 |
=1-
1 |
2 |
1 |
2 |
1 |
4 |
1 |
4 |
1 |
8 |
1 |
128 |
1 |
256 |
=1-
1 |
256 |
=
255 |
256 |
(3)
1 |
1×4 |
1 |
4×7 |
1 |
7×10 |
1 |
97×100 |
=
1 |
3 |
1 |
4 |
1 |
4 |
1 |
7 |
1 |
7 |
1 |
10 |
1 |
97 |
1 |
100 |
=
1 |
3 |
1 |
100 |
=
1 |
3 |
99 |
100 |
=
33 |
100 |
(4)2011×20122012-2012×20112011,
=2011×2012×10001-2012×2011×10001,
=0;
(5)(5
5 |
9 |
4 |
9 |
4 |
5 |
2 |
5 |
=(8-0.8)×(7.6×1.25+2.4×1.25),
=7.2×[(7.6+2.4)×1.25],
=7.2×12.5,
=0.9×(8×12.5),
=0.9×100,
=90;
(6)
1×3×24+2×6×48+3×9×72 |
1×2×4+2×4×8+3×6×12 |
=
1×3×24×(1+8+27) |
1×2×4×(1+8+27) |
=9;
(7)设1+
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
1 |
2 |
1 |
3 |
1 |
4 |
(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 |
=a×(b+
1 |
5 |
1 |
5 |
=ab+
1 |
5 |
1 |
5 |
=
1 |
5 |
=
1 |
5 |
=
1 |
5 |
点评:此题考查了四则运算的简便计算,注意分析式中数据,灵活运用运算定律或运算技巧,进行简便计算.
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