考點:整式的加減
專題:
分析:(1)直接找出同類項,進而合并得出即可;
(2)首先去括號,進而找出同類項,合并同類項即可;
(3)首先去括號,進而找出同類項,合并同類項即可;
(4)首先去括號,進而找出同類項,合并同類項即可;
(5)首先去括號,進而找出同類項,合并同類項即可;
(6)首先去括號,進而找出同類項,合并同類項即可;
(7)首先去括號,進而找出同類項,合并同類項即可;
(8)首先去括號,進而找出同類項,合并同類項即可.
解答:解:(1)5x
2-8x+5-3x
2+6x=2x
2-2x+5;
(2)a
3b+(a
3b-2c)-2(a
3b-c)
=a
3b+a
3b-2c-2a
3b+2c,
=0;
(3)-3(2x
2-xy)+4(x
2-xy-6)
=-6x
2+3xy+4x
2-4xy-24
=-2x
2-xy-24;
(4)2(2a
2-9b)+3(-5a
2-4b)-3b
=4a
2-18b-15a
2-12b-3b,
=-11a
2-33b;
(5)(8a
2-3ab-5b
2)-(2a
2-2ab+3b
2)
=8a
2-3ab-5b
2-2a
2+2ab-3b
2,
=6a
2-8b
2-ab;
(6)-4xy+3(
xy-2x)=-4xy+xy-6x=-3xy-6x;
(7)2a+2(a+1)-3(a-1)=2a+2a+2-3a+3=a+5;
(8)-3(2x
2-xy)+4(x
2+xy-6)=-6x
2+xy+4x
2+4xy-24=-2x
2+5xy-24.
點評:此題主要考查了整式的加減運算,正確去括號后合并同類項是解題關鍵.