试题
题目:
计算与化简
(1)
(
3
8
-
1
6
-
3
4
)×(-24)
(2)
-
1
4
-
1
6
×[3-
(-3)
2
]
(3)4(m
2
+n)+2(n-2m
2
) (4)5ab
2
-[a
2
b+2(a
2
b-3ab
2
)]
答案
解:(1)
(
3
8
-
1
6
-
3
4
)×(-24)
=-
3
8
×24+
1
6
×24+
3
4
×24
=-9+4+18
=13
(2)
-
1
4
-
1
6
×[3-
(-3)
2
]
=-1-
1
6
×(3-9)
=-1+1
=0
(3)4(m
2
+n)+2(n-2m
2
)
=4m
2
+4n+2n-4m
2
=6n
(4)5ab
2
-[a
2
b+2(a
2
b-3ab
2
)]
=5ab
2
-(a
2
b+2a
2
b-6ab
2
)
=5ab
2
-3a
2
b+6ab
2
=11ab
2
-3a
2
b
解:(1)
(
3
8
-
1
6
-
3
4
)×(-24)
=-
3
8
×24+
1
6
×24+
3
4
×24
=-9+4+18
=13
(2)
-
1
4
-
1
6
×[3-
(-3)
2
]
=-1-
1
6
×(3-9)
=-1+1
=0
(3)4(m
2
+n)+2(n-2m
2
)
=4m
2
+4n+2n-4m
2
=6n
(4)5ab
2
-[a
2
b+2(a
2
b-3ab
2
)]
=5ab
2
-(a
2
b+2a
2
b-6ab
2
)
=5ab
2
-3a
2
b+6ab
2
=11ab
2
-3a
2
b
考点梳理
考点
分析
点评
专题
整式的加减;有理数的混合运算.
(1)运用分配律展开计算即可;
(2)计算时注意运算顺序,先算乘方,再算乘除,最后算加减;
(3)去括号展开后合并同类项即可;
(4)去括号后合并同类项.
本题考查了有理数和整式的混合运算,解题的关键是正确的运用运算律和运算法则,特别要注意符号.
计算题.
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