试题

题目:
先化简,再求值.
(1)(5a3+3)-(1-2a)+3(3a-a3),其中a=-1.
(2)2x2y-[3xy2+2(xy2+2x2y)],其中x=
1
2
,y=-2.
(3)已知A=5x2+4x-1,B=-x2-3x+3,C=8-7x-6x2,求A-B+C的值.
答案
解:(1)∵a=-1,
∴原式=5a3+3-1+2a+9a-3a3
=2a3+11a+2
=-2-11+2
=-11,

(2)∵原式=2x2y-[3xy2+2xy2+4x2y]
=2x2y-3xy2-2xy2-4x2y   
=-2x2y-5xy2
∴当x=
1
2
,y=-2时,
原式=-2x2y-5xy2
=-2×(
1
2
)
2
×(-2)-5×
1
2
×(-2)2
 
=-9,
                        
(3)∵A=5x2+4x-1,B=-x2-3x+3,C=8-7x-6x2
∴A-B+C=(5x2+4x-1)-(-x2-3x+3)+(8-7x-6x2) 
=5x2+4x-1+x2+3x-3+8-7x-6x2 
=4.
解:(1)∵a=-1,
∴原式=5a3+3-1+2a+9a-3a3
=2a3+11a+2
=-2-11+2
=-11,

(2)∵原式=2x2y-[3xy2+2xy2+4x2y]
=2x2y-3xy2-2xy2-4x2y   
=-2x2y-5xy2
∴当x=
1
2
,y=-2时,
原式=-2x2y-5xy2
=-2×(
1
2
)
2
×(-2)-5×
1
2
×(-2)2
 
=-9,
                        
(3)∵A=5x2+4x-1,B=-x2-3x+3,C=8-7x-6x2
∴A-B+C=(5x2+4x-1)-(-x2-3x+3)+(8-7x-6x2) 
=5x2+4x-1+x2+3x-3+8-7x-6x2 
=4.
考点梳理
整式的加减—化简求值;合并同类项;去括号与添括号.
(1)首先进行乘法运算,再根据去括号法则去掉括号、合并同类项,然后把a的值代入求值即可,(2)首先根据去括号法则去掉小括号,对中括号内的代数式合并同类项,然后去掉中括号,再进行合并同类项,最后把x、y的值代入求值即可,(3)首先把A,B,C所表示的代数式代入到A-B+C,然后合并同类项即可.
本题主要考查整式的化简求值,合并同类先、去括号法则等知识点,关键在于正确的去括号、合并同类项,认真的进行计算.
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