题目内容
(1)
÷[1÷(1.6-1.599)]
(2)[
÷
-(
+
)]÷
(3)99×
(4)1-
+
-
+
-
+
-
.
| 1 |
| 32 |
(2)[
| 5 |
| 4 |
| 4 |
| 3 |
| 3 |
| 8 |
| 1 |
| 4 |
| 15 |
| 32 |
(3)99×
| 97 |
| 98 |
(4)1-
| 5 |
| 6 |
| 7 |
| 12 |
| 9 |
| 20 |
| 11 |
| 30 |
| 13 |
| 42 |
| 15 |
| 56 |
| 17 |
| 72 |
分析:(1)先算小括号里面的减法,再算中括号里面的除法,最后算括号外的除法;
(2)先算小括号里面的加法,再算中括号里面的除法,然后算中括号里面的减法,最后算括号外的除法;
(3)先把99分解成98+1,再运用乘法分配律简算;
(4)通过观察发现这些分数有一定的特点,分子与分子、分母与分母都按一定的规律递增,并且每一个分数都能拆成两个分数相加的形式,所以我们就进行分数的拆项,拆项后,通过前后两个分数相互抵消,达到简算的目的.
(2)先算小括号里面的加法,再算中括号里面的除法,然后算中括号里面的减法,最后算括号外的除法;
(3)先把99分解成98+1,再运用乘法分配律简算;
(4)通过观察发现这些分数有一定的特点,分子与分子、分母与分母都按一定的规律递增,并且每一个分数都能拆成两个分数相加的形式,所以我们就进行分数的拆项,拆项后,通过前后两个分数相互抵消,达到简算的目的.
解答:解:(1)
÷[1÷(1.6-1.599)],
=
÷[1÷0.001],
=
÷1000,
=
;
(2)[
÷
-(
+
)]÷
,
=[
÷
-
]÷
,
=[
-
]÷
,
=
×
,
=
;
(3)99×
,
=(98+1)×
,
=98×
+1×
,
=97+
,
=97
;
(4)1-
+
-
+
-
+
-
,
=1-(
+
)+(
+
)-(
-
)-…-(
+
),
=1-
-
+
+
-
-
-…-
-
,
=1-
-
,
=
-
,
=
.
| 1 |
| 32 |
=
| 1 |
| 32 |
=
| 1 |
| 32 |
=
| 1 |
| 32000 |
(2)[
| 5 |
| 4 |
| 4 |
| 3 |
| 3 |
| 8 |
| 1 |
| 4 |
| 15 |
| 32 |
=[
| 5 |
| 4 |
| 4 |
| 3 |
| 5 |
| 8 |
| 15 |
| 32 |
=[
| 15 |
| 16 |
| 5 |
| 8 |
| 15 |
| 32 |
=
| 5 |
| 16 |
| 32 |
| 15 |
=
| 2 |
| 3 |
(3)99×
| 97 |
| 98 |
=(98+1)×
| 97 |
| 98 |
=98×
| 97 |
| 98 |
| 97 |
| 98 |
=97+
| 97 |
| 98 |
=97
| 97 |
| 98 |
(4)1-
| 5 |
| 6 |
| 7 |
| 12 |
| 9 |
| 20 |
| 11 |
| 30 |
| 13 |
| 42 |
| 15 |
| 56 |
| 17 |
| 72 |
=1-(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 8 |
| 1 |
| 9 |
=1-
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 8 |
| 1 |
| 9 |
=1-
| 1 |
| 2 |
| 1 |
| 9 |
=
| 1 |
| 2 |
| 1 |
| 9 |
=
| 7 |
| 18 |
点评:本题考查了四则混合运算,注意运算顺序和运算法则,灵活运用所学的运算定律进行简便计算.
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