题目内容
(2013?广州模拟)计算下列各题,能简便的要简便计算
3
÷(1+
×1
)
(
+
)×15×17
(9.6-10
÷1.25)÷(1
×0.75)
5.66×7.28-3.16×3.16-3.16×4.12
2012×0.875+201.2×1.25
+
+
+…+
.
3
1 |
6 |
1 |
3 |
3 |
4 |
(
1 |
15 |
2 |
17 |
(9.6-10
1 |
2 |
1 |
3 |
5.66×7.28-3.16×3.16-3.16×4.12
2012×0.875+201.2×1.25
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
99×100 |
分析:(1)按运算顺序进行计算,先算括号内的,再算括号外的;
(2)根据数字特点,运用乘法分配律简算;
(3)按运算顺序进行计算,两个括号同时运算,第一个括号内既有除法运算,也有减法运算,应先算除法,再算减法;
(4)根据数字特点,先运用乘法分配律的逆运算,把原式改写为5.66×7.28-3.16×(3.16+4.12),然后再运用乘法分配律的逆运算简算;
(5)根据积不变的规律,把201.2×1.25改写成2012×0.125,然后运用乘法分配律的逆运算计算;
(6)通过观察,分母是两个连续自然数的乘积,可以把每个分数改写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)根据数字特点,运用乘法分配律简算;
(3)按运算顺序进行计算,两个括号同时运算,第一个括号内既有除法运算,也有减法运算,应先算除法,再算减法;
(4)根据数字特点,先运用乘法分配律的逆运算,把原式改写为5.66×7.28-3.16×(3.16+4.12),然后再运用乘法分配律的逆运算简算;
(5)根据积不变的规律,把201.2×1.25改写成2012×0.125,然后运用乘法分配律的逆运算计算;
(6)通过观察,分母是两个连续自然数的乘积,可以把每个分数改写成两个分数相减的形式,然后通过加、减相互抵消,求出结果.
解答:解:(1)3
÷(1+
×1
),
=3
÷(1+
×
),
=3
÷(1+
),
=
÷
,
=
×
,
=2;
(2)(
+
)×15×17,
=
×15×17+
×15×17,
=17+30,
=47;
(3)(9.6-10
÷1.25)÷(1
×0.75),
=(9.6-
÷
)÷(
×
),
=(9.6-
×
)÷1,
=(9.6-8.4)÷1,
=1.2;
(4)5.66×7.28-3.16×3.16-3.16×4.12,
=5.66×7.28-3.16×(3.16+4.12),
=5.66×7.28-3.16×7.28,
=(5.66-3.16)×7.28,
=2.5×7.28,
=18.2;
(5)2012×0.875+201.2×1.25,
=2012×0.875+2012×0.125,
=2012×(0.875+0.125),
=2012×1,
=2012;
(6)
+
+
+…+
,
=1-
+
-
+
-
+…+
-
,
=1-
,
=
.
1 |
6 |
1 |
3 |
3 |
4 |
=3
1 |
6 |
1 |
3 |
7 |
4 |
=3
1 |
6 |
7 |
12 |
=
19 |
6 |
19 |
12 |
=
19 |
6 |
12 |
19 |
=2;
(2)(
1 |
15 |
2 |
17 |
=
1 |
15 |
2 |
17 |
=17+30,
=47;
(3)(9.6-10
1 |
2 |
1 |
3 |
=(9.6-
21 |
2 |
5 |
4 |
4 |
3 |
3 |
4 |
=(9.6-
21 |
2 |
4 |
5 |
=(9.6-8.4)÷1,
=1.2;
(4)5.66×7.28-3.16×3.16-3.16×4.12,
=5.66×7.28-3.16×(3.16+4.12),
=5.66×7.28-3.16×7.28,
=(5.66-3.16)×7.28,
=2.5×7.28,
=18.2;
(5)2012×0.875+201.2×1.25,
=2012×0.875+2012×0.125,
=2012×(0.875+0.125),
=2012×1,
=2012;
(6)
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
99×100 |
=1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
99 |
1 |
100 |
=1-
1 |
100 |
=
99 |
100 |
点评:此题考查了学生对四则混合运算顺序的掌握情况,以及认真审题、灵活解答的能力.
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