文档介绍：Forpersonaluseonlyinstudyandresearch;mercialuse芈肆罿整式训练专题训练螄蚀蚈1归纳出去括号的法则吗?:薄螁膄(1)a+(-b+c-d);(2)a-(-b+c-d);蝿芈肀(3)-(p+q)+(m-n);(4)(r+s)-(p-q).?若有错,请改正:袇蚈袅(1)a2-(2a-b+c)(2)-(x-y)+(xy-1)肅薀袄=a2-2a-b+c;=-x-y+xy-(3)(y-x)2=(x-y)2螅蚁聿(4)(-y-x)2=(x+y)2莈蒇莅(5)(y-x)3=(x-y):螀羆衿(1)(2x-3y)+(5x+4y);(2)(8a-7b)-(4a-5b);芆蒀芇(3)a-(2a+b)+2(a-2b);(4)3(5x+4)-(3x-5);衿莅螄(5)(8x-3y)-(4x+3y-z)+2z;(6)-5x2+(5x-8x2)-(-12x2+4x)+2;螂薁莅(7)2-(1+x)+(1+x+x2-x2);(8)3a2+a2-(2a2-2a)+(3a-a2)。羇螅羀作业:,在___上填上“+”号或“-”号:荿蒈蒈(1)a___(-b+c)=a-b+c;(2)a___(b-c-d)=a-b+c+d;芃莀袁(3)___(a-b)___(c+d)=c+d-a++y=2,则x+y+3= ,5-x-y= .:螀莇袇(1)a+3(2b+c-d); (2)3x-2(3y+2z).蚄薃薂(3)3a+4b-(2b+4a); (4)(2x-3y)-3(4x-2y).:蒄莀袇(1)2a-3b+[4a-(3a-b)];(2)3b-2c-[-4a+(c+3b)]+:-[2(x+3y)-3(x-2y)]的结果是().+2; -12y+2; C.-5x+12y+2; -:+=3,求{x-[x2-(1-x)]}-,与a-b-c的值不相等的是()-(b+c)-(b-c)袀螈羂C.(a-b)+(-c)D.(-c)+(-b+a)-[0-(2p-q)]的结果是()羈***膅A.-2p-qB.-2p+-+,正确的是()-(2b-3c)=a-2b--(3x2+2x-1)=x3-3x2-2x-+(-2y+1)=2y2-2y+1莃节薃D.-(2x-y)-(-x2+y2)=-2x+y+x2+:膃芃螈a+(b-c)=;(a-b)+(-c-d)=;羀袄芈-(a-b)-(-c-d)=;袃肁莄5x3-[3x2-(x-1)]=.(1)x-(y-z)=x-y-z()蒁羇蚇(2)-(x-y+z)=-x+y-z()莄罿肄(3)x-2(y-z)=x-2y+z()蕿蒇虿(4)-(a-b)+(-c-d)=-a+b+c+d():蚇袆***-(2m-3);n-3(4-2m);袅肂螅(1)16a-8(3b+4c);(2)-(x+y)+(p+q);肀芅蚁(3)-8(3a-2ab+4);(4)4(rn+p)-7(n-2q).薅袀莇(5)8(y-x)2-(x-y)2-4(-y-x)2-3(x+y)2+2(y-x),再合并同类项:莆袁薅-2n-(3n-1);a-(5a-3b)+(2b-a);薀莈螂-3(2s-5)+6s;1-(2a-1)-(3a+3);袂羂螀3(-ab+2a)-(3a-b);14(abc-2a)+3(6a-2abc).-︱-[a-(b-c)]︱去括号后的结果应为()+b+-.-a+b--b-(3-)-︱-3︱的结果为().---,合并同类项;肃袂螂6a2-2ab-2(3a2-ab);2(2a-b)-[4b-(-2a+b)]芈肆薁9a3-[-6a2+2(a3-a2)];2t-[t-(t2-t-3)-2]+(2t2-3t+1).,并求出代数式25+3a-{11a-[a-10-7(1-a)]}的值,你能从中发现什么?:(1)102+199-99;(2)5040-297-1503的简便方法计算芄螃荿解:(1)102+199-99(2)5040-297-1503袇蚈薇=102+(199-99)=5040-(297+1503)肅薀节=102+10