试题

题目:
计算
(1)(8a-6b)-(4a-5b)+(3a-2b)   
(2)3x2y-2[
1
2
x2y-(x2y-
1
4
x2)-2x2]

(3)先化简再求值:2a2-[
1
2
(ab-4a2)+8ab]-
1
2
ab
,其中a=-
1
2
,b=
2
3

答案
解:(1)原式=8a-6b-4a+5b+3a-2b
=(8a-4a+3a)+(5b-6b-2b)
=7a-3b;

(2)原式=3x2y-2(
1
2
x2y-x2y+
1
4
x2-2x2
=3x2y-x2y+2x2y-
1
2
x2+4x2
=4x2y+
7
2
x2

(3)原式=2a2-(
1
2
ab-2a2+8ab)-
1
2
ab
=2a2-
1
2
ab+2a2-8ab-
1
2
ab
=4a2-9ab,
当a=-
1
2
,b=
2
3
时,
原式=4×( -
1
2
)
2
-9×(-
1
2
)×
2
3

=1+3
=4.
解:(1)原式=8a-6b-4a+5b+3a-2b
=(8a-4a+3a)+(5b-6b-2b)
=7a-3b;

(2)原式=3x2y-2(
1
2
x2y-x2y+
1
4
x2-2x2
=3x2y-x2y+2x2y-
1
2
x2+4x2
=4x2y+
7
2
x2

(3)原式=2a2-(
1
2
ab-2a2+8ab)-
1
2
ab
=2a2-
1
2
ab+2a2-8ab-
1
2
ab
=4a2-9ab,
当a=-
1
2
,b=
2
3
时,
原式=4×( -
1
2
)
2
-9×(-
1
2
)×
2
3

=1+3
=4.
考点梳理
整式的加减—化简求值;整式的加减.
(1)去括号,合并同类项即可;
(2)去括号,合并同类项,注意符号;
(2)首先去括号合并同类项,再代入求值.
此题考查的知识点是整式的加减-化简求值,关键是去括号合并同类项,且注意符号.
计算题.
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