试题

题目:
化简
(1)2x2y-2xy-4xy2+xy+4x2y-3xy2&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;
(2)3&nb5p;(4x2-3x+2)-2&nb5p;(1-4x2+x)
(3)5abc-2a2b-[3abc-3&nb5p;(4ab2+a2b)]
(4)(2x2+x)-2[x2-2(3x2-x)].
答案
解:(v)2x2y-2xy-4xy2+xy+4x2y-3xy2=2x2y+4x2y-2xy+xy-4xy2-3xy2=6x2y-xy-7xy2
(2)3&nbs七;(4x2-3x+2)-2&nbs七;(v-4x2+x)=v2x2-9x+6-2+8x2-2x=2dx2-vvx+4
(3)5abc-2a2b-[3abc-3&nbs七;(4ab2+a2b)]=5abc-2a2b-[3abc-v2ab2-3a2b]=5abc-2a2b-3abc+v2ab2+3a2b=2abc+a2b+v2ab2
(4)(2x2+x)-2[x2-2(3x2-x)]=2x2+x-2[x2-6x2+2x]=2x2+x-2x2+v2x2-4x=v2x2-3x;
解:(v)2x2y-2xy-4xy2+xy+4x2y-3xy2=2x2y+4x2y-2xy+xy-4xy2-3xy2=6x2y-xy-7xy2
(2)3&nbs七;(4x2-3x+2)-2&nbs七;(v-4x2+x)=v2x2-9x+6-2+8x2-2x=2dx2-vvx+4
(3)5abc-2a2b-[3abc-3&nbs七;(4ab2+a2b)]=5abc-2a2b-[3abc-v2ab2-3a2b]=5abc-2a2b-3abc+v2ab2+3a2b=2abc+a2b+v2ab2
(4)(2x2+x)-2[x2-2(3x2-x)]=2x2+x-2[x2-6x2+2x]=2x2+x-2x2+v2x2-4x=v2x2-3x;
考点梳理
整式的加减;合并同类项;去括号与添括号.
(1)按照有理数混合运算的顺序,先把同类项放在一起,再合并,即可求出结果;
(2)先去括号,然后合并同类项.注意去括号时,如果括号前是负号,那么括号中的每一项都要变号;合并同类项时,只把系数相加减,字母与字母的指数不变.
(3)先去小括号,再去中括号,然后合并同类项即可求出答案;
(4)先去小括号,再去中括号,然后合并同类项即可求出答案;
本题考查的是有理数的运算与整式的加减运算.注意:要正确掌握运算顺序,即乘方运算(和以后学习的开方运算)叫做三级运算;乘法和除法叫做二级运算;加法和减法叫做一级运算.在混合运算中要特别注意运算顺序:先三级,后二级,再一级;有括号的先算括号里面的;同级运算按从左到右的顺序.
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