试题

题目:
先化简再求值.
3x2y-[2xy2-2(xy-
3
2
x2y)+xy]+3xy2,其中x=3,y=-
1
3

②(2a+3b-2ab)-(a+4b+ab)-(3ab-2a+2b),其中a-b=5,ab=-5
答案
解:①原式=3x2y-(2xy2-2xy+3x2y+xy)+3xy2
=3x2y-2xy2+2xy-3x2y-xy+3xy2
=xy2+xy
将x=3,y=-
1
3
代入上式,得
上式=3×(-
1
3
)2
+3×(-
1
3
)

=3×
1
9
-1
=
1
3
-1

=-
2
3

②(2a+3b-2ab)-(a+4b+ab)-(3ab-2a+2b)
=2a+3b-2ab-a-4b-ab-3ab+2a-2b
=3a-3b-6ab
=3(a-b)-6ab
将a-b=5,ab=-5代入上式,得
上式=3×5-6×(-5)
=15+30
=45.
解:①原式=3x2y-(2xy2-2xy+3x2y+xy)+3xy2
=3x2y-2xy2+2xy-3x2y-xy+3xy2
=xy2+xy
将x=3,y=-
1
3
代入上式,得
上式=3×(-
1
3
)2
+3×(-
1
3
)

=3×
1
9
-1
=
1
3
-1

=-
2
3

②(2a+3b-2ab)-(a+4b+ab)-(3ab-2a+2b)
=2a+3b-2ab-a-4b-ab-3ab+2a-2b
=3a-3b-6ab
=3(a-b)-6ab
将a-b=5,ab=-5代入上式,得
上式=3×5-6×(-5)
=15+30
=45.
考点梳理
整式的加减—化简求值.
把①②先去括号,再合并同类项,然后将已知条件代入求值.
合并同类项的法则:同类项的系数相加,所得结果作为系数,字母和字母是指数不变.
找相似题