题目内容
耐心算一算
(1)(-3)+(-4)-(+11)-(-19)
(2)-23-(1-0.5)×
×[2-(-3)2]
(3)-3.5÷
×(-
)×|-
|
(4)(
-
-
)×(-60)
(5)(-27
)÷9(用简便方法计算)
(6)(4x2y-3xy2)-(1+4x2y-3xy2)
(1)(-3)+(-4)-(+11)-(-19)
(2)-23-(1-0.5)×
| 1 |
| 3 |
(3)-3.5÷
| 7 |
| 8 |
| 8 |
| 7 |
| 3 |
| 64 |
(4)(
| 2 |
| 3 |
| 1 |
| 12 |
| 4 |
| 15 |
(5)(-27
| 9 |
| 11 |
(6)(4x2y-3xy2)-(1+4x2y-3xy2)
分析:(1)把减法运算转化为加法运算得到原式=-3-4-11+19,然后根据有理数加法法则进行运算即可;
(2)先算乘方和括号得到原式=-8-
×
×(2-9)=-8-
×(-7),再进行乘法运算得到原式=-8+
,然后根据有理数加法法则进行运算即可;
(3)先把小数化为分数和去绝对值得到原式=-
×
×(-
)×
,然后进行乘法运算即可;
(4)根据乘法的分配律得到原式=
×(-60)-
×(-60)-
×(-60),再进行乘法运算得到原式=-40+5+16,然后根据有理数加法法则进行运算即可;
(5)把带分数化为(27+
),再根据乘法的分配律得到原式=-(27×
+
×
),再进行乘法运算得到原式=-(3+
),然后进行加法运算即可;
(6)去括号得到原式=4x2y-3xy2-1-4x2y+3xy2,然后合并同类项即可.
(2)先算乘方和括号得到原式=-8-
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 6 |
| 7 |
| 6 |
(3)先把小数化为分数和去绝对值得到原式=-
| 7 |
| 2 |
| 8 |
| 7 |
| 8 |
| 7 |
| 3 |
| 64 |
(4)根据乘法的分配律得到原式=
| 2 |
| 3 |
| 1 |
| 12 |
| 4 |
| 15 |
(5)把带分数化为(27+
| 9 |
| 11 |
| 1 |
| 9 |
| 9 |
| 11 |
| 1 |
| 9 |
| 1 |
| 11 |
(6)去括号得到原式=4x2y-3xy2-1-4x2y+3xy2,然后合并同类项即可.
解答:(1)解:原式=-3-4-11+19
=-18+19
=1;
(2)解:原式=-8-
×
×(2-9)
=-8-
×(-7)
=-8+
=-
;
(3)解:原式=-
×
×(-
)×
=
;
(4)解:原式=
×(-60)-
×(-60)-
×(-60)
=-40+5+16
=-19;
(5)解:原式=-(27+
)×
=-(27×
+
×
)
=-(3+
)
=-
;
(6)解:原式=4x2y-3xy2-1-4x2y+3xy2
=-1.
=-18+19
=1;
(2)解:原式=-8-
| 1 |
| 2 |
| 1 |
| 3 |
=-8-
| 1 |
| 6 |
=-8+
| 7 |
| 6 |
=-
| 41 |
| 6 |
(3)解:原式=-
| 7 |
| 2 |
| 8 |
| 7 |
| 8 |
| 7 |
| 3 |
| 64 |
=
| 3 |
| 14 |
(4)解:原式=
| 2 |
| 3 |
| 1 |
| 12 |
| 4 |
| 15 |
=-40+5+16
=-19;
(5)解:原式=-(27+
| 9 |
| 11 |
| 1 |
| 9 |
=-(27×
| 1 |
| 9 |
| 9 |
| 11 |
| 1 |
| 9 |
=-(3+
| 1 |
| 11 |
=-
| 34 |
| 11 |
(6)解:原式=4x2y-3xy2-1-4x2y+3xy2
=-1.
点评:本题考查了有理数的混合运算:先算乘方,再进行有理数的乘除运算,最后进行有理数的加减运算;有括号先计算括号;利用运算律可简化运算.也考查了合并同类项.
练习册系列答案
相关题目