.(2a)³·b⁴÷12a³b²

|2a-b+1|+(3a+3/2b)^2=0绝对值和平方都大于等于0现在相加等于0若有一个大于0,则另一个小于0,不可能所以两个都等于0所以2a-b+1=03a+3b/2=0b=2a+13a+3a+3

{[(2a-b)/(a+b)]-[b/(a-b)]}÷[(a-2b)/(a+b)]=(2a²-4ab)/[(a+b)(a-b)]×(a+b)/(a-2b)=(2a²-4ab)/(a

=2b2/a+a=(2b2-a2)/a

=[(2a-b)(a-b)/(a+b)(a-b)-b(a+b)/(a+b)(a-b)]×(a+b)/(a-2b)=(2a²-3ab+b²-ab-b²)/(a+b)(a-b

分母分子都乘以(a+b)(a-b)(2a^2-3ab+b^2-ab-b^2)/(a^2-3ab-2b^2)(2a^2-4ab)/((a-2b)(a-b))2a(a-2b)/((a-2b)(a-b))2

[（a+b)(a-b)-（a-b)的二次方-2b(b-a)]÷（4b)=[(a+b)(a-b)-(a-b)²+2b(a-b)]÷4b=(a-b)(a+b-a+b+2b)÷4b=a-b=-(b

：((a+b)(a-b)-(a-b)^2+2b(a-b))/4(a-b)=(a-b)[(a+b)-(a-b)+2b]/4(a-b)=(a-b)[2b+2b]/4(a-b)=

a/b(a-b)-b/a(a-b)；=[a²(a-b)-b²(a-b)]/ab(a-b)=(a-b)²(a+b)/ab(a-b)=(a²-b²)/ab

a-b-1a-b分之（a+b)²再问：过程再答：a²-2ab+b²-a+b要分组，a²-2ab+b²一组等于（a-b)²；-a+b=-（a-

(b/a)^2÷（a/b)-b^3/a^3(a-b)÷1/(a-b)=(b/a)^2×（b/a)-b^3×(a-b)/a^3(a-b)=(b/a)^3-b^3/a^3=(b/a)^3-(b/a）^3＝

2a-b/a+b-b/a-b)除以a-2b/a+b=[(2a-b)(a-b)-b(a+b)]/(a+b)(a-b)x(a+b)/(a-2b)=(2a²-3ab+b²-ab-b

原式=(b-a)(b+a)/a(a-b)÷（a+b)²/a·(a+b)/(ab)=-1/(a+b)·(a+b)/(ab)=-1/(ab)

解原式=(a+b)(a^2+2ab+b^2-2a-2b-4)/(a+b)=a^2+2ab+b^2-2a-2b-4再问：求详细过程再答：解原式=(a+b)[(a+b)^2-2(a+b)-4]/(a+b)

解题思路：本题考查整式的除法，注意转化为乘法后，能约分的要约分，化为最简形式解题过程：

先化简,再求值;［(2a-b/a+b）-（b/a-b)］÷{（a-2b/a+b）+［ab/(a-b)²］},其中b/a=-1/2解,得：［(2a-b)(a-b)-b(a+b)]/[(a+b)

(a²+b²/a²-b²-a-b/a+b)÷2ab/(a-b)(a+b)²=(a²+b²/a²-b²-a-b/

(2a-b/a+b-b/a-b)除以a-2b/a+b=[(2a-b)(a-b)-b(a+b)]/(a+b)(a-b)x(a+b)/(a-2b)=(2a²-3ab+b²-ab-b&#

由a(a-1)-(a×a-b)=-5得a^2-a-a^2+b=-5,即a-b=5,(a-b)^2=25.(a×a+b×b)÷2-ab=(a-b)^2÷2=25÷2=12.5

(a-b)/(2a+2b)·(a^2+b^2)/(a^2-b^2)=(a-b)/2(a+b)×(a²+b²)/(a+b)(a-b)=(a²+b²)/2(a+b)

[(a+2b)(a-2b)-(2a-b)^2+(3a-b)(3a-5b)]÷2a=(a²-4b²-4a²+4ab-b²+9a²-18ab+5b