题目内容
去括号且合并后,把结果写在右边的横线上.
(1)32b-(2b-5)=
(2)
x+
(2x-1)=
x-
x-
;
(3)4y-2(3y+
)=
(4)2a-
(a+3)=
a-2
a-2;
(5)1.5y+2-2(10-0.5y)=
(6)2x-[-3(2-6x)]=
(1)32b-(2b-5)=
30b+5
30b+5
;(2)
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(3)4y-2(3y+
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-2y-
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-2y-
;| 1 |
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(4)2a-
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(5)1.5y+2-2(10-0.5y)=
2.5y-18
2.5y-18
;(6)2x-[-3(2-6x)]=
-16x+6
-16x+6
.分析:根据去括号(添括号),括号前是负号,去括号(添括号)要变号,括号前是正号,去括号(添括号)不变号,可得结果.
解答:解:(1)32b-(2b-5)
=32b-2b+5=30b+5;
(2)
x+
(2x-1)
=
x+
x-
=
x-
;
(3)4y-2(3y+
)
=4y-6y-
=-2y-
;
(4)2a-
(a+3)
=2a-
a-2=
a-2;
(5)1.5y+2-2(10-0.5y)
=1.5y+2-20+y=2.5y-18;
(6)2x-[-3(2-6x)]
=2x-(-6+18x)=2x+6-18x=-16x+6;
故答案为:30b+5,
x-
,-2y-
,
a-2,2.5y-18,-16x+6.
=32b-2b+5=30b+5;
(2)
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=
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(3)4y-2(3y+
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=4y-6y-
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(4)2a-
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| 3 |
=2a-
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(5)1.5y+2-2(10-0.5y)
=1.5y+2-20+y=2.5y-18;
(6)2x-[-3(2-6x)]
=2x-(-6+18x)=2x+6-18x=-16x+6;
故答案为:30b+5,
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点评:本题考查了去括号与添括号,根据法则去括号(添括号),再合并同类项.
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