题目内容
计算
①243×3.75+24
×41.1+2
×214
②[(5
-
)÷2.5]×[1÷(2
-2
)]
③[(
+1
)×
-0.75]÷
④1+2-3-4+5+6-7-8+…+1997+1998.
①243×3.75+24
| 3 |
| 10 |
| 43 |
| 100 |
②[(5
| 4 |
| 5 |
| 4 |
| 5 |
| 1 |
| 10 |
| 9 |
| 100 |
③[(
| 8 |
| 9 |
| 1 |
| 3 |
| 3 |
| 4 |
| 1 |
| 12 |
④1+2-3-4+5+6-7-8+…+1997+1998.
考点:整数、分数、小数、百分数四则混合运算
专题:运算顺序及法则
分析:(1)先化24
×41.1+2
×214=243×4.11+243×2.14,再运用乘法分配律解答,
(2)按照先同时算小括号里面的,再同时算中括号里面的,最后算括号外面的顺序计算解答,
(3)按照先算小括号里面的加法,再算中括号里面的乘法,然后算中括号里面的减法,最后算括号外面的除法顺序计算解答,
(4)运用高斯定律解答以及乘法分配律解答.
| 3 |
| 10 |
| 43 |
| 100 |
(2)按照先同时算小括号里面的,再同时算中括号里面的,最后算括号外面的顺序计算解答,
(3)按照先算小括号里面的加法,再算中括号里面的乘法,然后算中括号里面的减法,最后算括号外面的除法顺序计算解答,
(4)运用高斯定律解答以及乘法分配律解答.
解答:
解:(1)243×3.75+24
×41.1+2
×214
=243×3.75+243×4.11+243×2.14
=243×(3.75+4.11+2.14)
=243×10
=2430;
(2)[(5
-
)÷2.5]×[1÷(2
-2
)]
=[5÷2.5]×[1÷
]
=2×100
=200;
(3)[(
+1
)×
-0.75]÷
=[
×
-0.75]÷
=[
-0.75]÷
=
÷
=11;
(4)1+2+3+4+5+6+7+8+…+1997+1998
=(1+1998)+(2+1997)+…+(999+1000)
=1999×999
=1999×1000-1999×1
=1999000-1999
=19970001.
| 3 |
| 10 |
| 43 |
| 100 |
=243×3.75+243×4.11+243×2.14
=243×(3.75+4.11+2.14)
=243×10
=2430;
(2)[(5
| 4 |
| 5 |
| 4 |
| 5 |
| 1 |
| 10 |
| 9 |
| 100 |
=[5÷2.5]×[1÷
| 1 |
| 100 |
=2×100
=200;
(3)[(
| 8 |
| 9 |
| 1 |
| 3 |
| 3 |
| 4 |
| 1 |
| 12 |
=[
| 20 |
| 9 |
| 3 |
| 4 |
| 1 |
| 12 |
=[
| 5 |
| 3 |
| 1 |
| 12 |
=
| 11 |
| 12 |
| 1 |
| 12 |
=11;
(4)1+2+3+4+5+6+7+8+…+1997+1998
=(1+1998)+(2+1997)+…+(999+1000)
=1999×999
=1999×1000-1999×1
=1999000-1999
=19970001.
点评:依据各数不同的特点,正确选择并运用不同方法解决问题,是此类题目考查知识点.
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