题目内容
计算下列各题:
(1)-20+(-14)-(-18)-13;
(2)10+(-2)×(-5)2;
(3)
÷
-
×(-6);
(4)(-
-
+
)÷(-
);
(5)|-
|÷(
-
)-
×(-4)2;
(6)-22-(-3)3×(-1)4-(-1)5;
(7)
+
+
+…+
;
(8)-3
÷(-
).
(1)-20+(-14)-(-18)-13;
(2)10+(-2)×(-5)2;
(3)
| 7 |
| 4 |
| 7 |
| 8 |
| 2 |
| 3 |
(4)(-
| 3 |
| 4 |
| 5 |
| 9 |
| 7 |
| 12 |
| 1 |
| 36 |
(5)|-
| 7 |
| 9 |
| 2 |
| 3 |
| 1 |
| 5 |
| 1 |
| 3 |
(6)-22-(-3)3×(-1)4-(-1)5;
(7)
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 1999×2000 |
(8)-3
| 23 |
| 24 |
| 1 |
| 12 |
分析:(1)先去掉括号,然后按照有理数的加法运算方法进行计算即可;
(2)先算乘方,再算乘除,最后算加减;
(3)先算乘除,后算加减;
(4)先将除法转化成乘法,然后利用分配律进行计算即可;
(5)先去掉绝对值并计算乘方,再做乘除;
(6)先算乘方,再算乘除,最后算加减即可;
(7)将原式转化为1-
+
-
+
-…+
+
即可求解;
(8)将原式转化为4+
)×12即可求解.
(2)先算乘方,再算乘除,最后算加减;
(3)先算乘除,后算加减;
(4)先将除法转化成乘法,然后利用分配律进行计算即可;
(5)先去掉绝对值并计算乘方,再做乘除;
(6)先算乘方,再算乘除,最后算加减即可;
(7)将原式转化为1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 1999 |
| 1 |
| 2000 |
(8)将原式转化为4+
| 1 |
| 24 |
解答:解:(1)-20+(-14)-(-18)-13
=-20-14+18-13
=-47+1=-29;
(2)10+(-2)×(-5)2;
=10+(-2)×25
=10-50
=-40;
(3)
÷
-
×(-6)
=
×
+4
=2+4
=6
(4)(-
-
+
)÷(-
)
=(-
-
+
)×(-36)
=
×36+
×36-
×36
=27+20-21
=26;
(5)|-
|÷(
-
)-
×(-4)2;
=
÷
-
×16
=
-
=-
(6)-22-(-3)3×(-1)4-(-1)5;
=-4+27+1
=24;
(7)
+
+
+…+
=1-
+
-
+
-…+
+
=1-
=
;
(8)-3
÷(-
)
=(-4+
)×12
=-48+
=47
.
=-20-14+18-13
=-47+1=-29;
(2)10+(-2)×(-5)2;
=10+(-2)×25
=10-50
=-40;
(3)
| 7 |
| 4 |
| 7 |
| 8 |
| 2 |
| 3 |
=
| 7 |
| 4 |
| 8 |
| 7 |
=2+4
=6
(4)(-
| 3 |
| 4 |
| 5 |
| 9 |
| 7 |
| 12 |
| 1 |
| 36 |
=(-
| 3 |
| 4 |
| 5 |
| 9 |
| 7 |
| 12 |
=
| 3 |
| 4 |
| 5 |
| 9 |
| 7 |
| 12 |
=27+20-21
=26;
(5)|-
| 7 |
| 9 |
| 2 |
| 3 |
| 1 |
| 5 |
| 1 |
| 3 |
=
| 7 |
| 9 |
| 7 |
| 15 |
| 1 |
| 3 |
=
| 15 |
| 9 |
| 16 |
| 3 |
=-
| 11 |
| 3 |
(6)-22-(-3)3×(-1)4-(-1)5;
=-4+27+1
=24;
(7)
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 1999×2000 |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 1999 |
| 1 |
| 2000 |
=1-
| 1 |
| 2000 |
=
| 1999 |
| 2000 |
(8)-3
| 23 |
| 24 |
| 1 |
| 12 |
=(-4+
| 1 |
| 24 |
=-48+
| 1 |
| 2 |
=47
| 1 |
| 2 |
点评:本题考查了有理数的混合运算,解题的关键是牢记有关法则的情况下认真的计算.
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