题目内容
先化简再求值.①3x2y-[2xy2-2(xy-
| 3 |
| 2 |
| 1 |
| 3 |
②(2a+3b-2ab)-(a+4b+ab)-(3ab-2a+2b),其中a-b=5,ab=-5
分析:把①②先去括号,再合并同类项,然后将已知条件代入求值.
解答:解:①原式=3x2y-(2xy2-2xy+3x2y+xy)+3xy2
=3x2y-2xy2+2xy-3x2y-xy+3xy2
=xy2+xy
将x=3,y=-
代入上式,得
上式=3×(-
)2+3×(-
)
=3×
-1
=
-1
=-
;
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.
=3x2y-2xy2+2xy-3x2y-xy+3xy2
=xy2+xy
将x=3,y=-
| 1 |
| 3 |
上式=3×(-
| 1 |
| 3 |
| 1 |
| 3 |
=3×
| 1 |
| 9 |
=
| 1 |
| 3 |
=-
| 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.
点评: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.
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