已知ab=4,a-b=3,求a的4次方减b的4次方

（2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)=3a-3b-6ab=3(a-b)-6ab=3*2-6*(-3)=6+18=24

ab+a=-2,ab+b=-10两等式相减,得：a-b=8-2a-{-4b-3[(ab+a)²-a]}-1/2[(ab+b)²-2b]=-2a-{-4b-3[(-2)²-

(-2ab+2a+3b)-(3ab+2b-2a)-(a+4b+ab)=-2ab+2a+3b-3ab-2b+2a-a-4b-ab=(-2ab-3ab-ab)+(2a+2a-a)+(3b-2b-4b)=-

此题考查的一点意义都没有.已知a+b=2,ab+b=-10,我们用a=2-b代替到另一个式子,即（2-b）b+b=-10,b平方-3b-10=0,这是和一元二次方程,即可求得：b=5或-2,那么对应a

2a+3ab+2b=2a+2b+3ab=2(a+b)+3ab=2*4+3*1=11a³-a²b+a²b²+a²b-a²b²+b&#

（3a-4b-ab)-(a-2b+3ab)=3a-4b-ab-a+2b-3ab=2a-2b-4ab=2（a-b）-4ab=2×（-1）-4×（-2）（将a-b=-1,ab=-2代入）=-2+8=6

(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)=2a+3b-2ab-a-4b-ab-3ab-2b+2a=3a-3b-6ab=3(a-b)-6ab已知a-b=3,ab=-33(a-b

原式=5a²b(a-b)-3ab(a-b)²-5ab²(a-b)=ab(a-b)[5a-3(a-b)-5b]=ab(a-b)(5a-3a+3b-5b)=ab(a-b)(2

解a²b+ab²-a³b²-a²b³=ab(a+b)-(ab)²(a+b)=-1/5×(4/5)-(-1/5)²×(4/

3a²b+3ab²=3ab（a+b）=3×3×4=36

1/a+1/b=7(a+b)/ab=7a+b=7ab原式=[(a+b)-4ab]/[2(a+b)-3ab]=(7ab-4ab)/(2×7ab-3ab)=3ab/11ab=3/11

(3a-4ab-3b)/(a-2ab-b)=（3（a-b）-4ab）/(a-b-2ab)=(3(1/b-1/a)-4)/(1/b-1/a-2)(这一步是分子分母同时除以ab)=（3x(-5)-4）/(

a-b=3,ab=-2（2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)=2a+3b-2ab-a-4b-ab-3ab-2b+2a=3a-3b-6ab=3*3-6*(-2)=21

(-a-4b-ab)-(2ab-2a-3b)-(3ab+2b-2a)=-a-4b-ab-2ab+2a+3b-3ab-2b+2a=3a-3b-6ab=3(a-b)-6ab=9+18=27

（22a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)=22a+3b-2ab-a-4b-ab-3ab-2b+2a=23a-3b-5ab题目抄差了!（22a+3b-2ab)应该是（2a+3

a+3ab-b=a-b+3ab=4+3×（-1）=1

由a-b=4,ab=-1解得b=-2+3^0.5或-2-3^0.5故有(-2ab+2a+3b)-(a-4b+b)=-2ab+a+6b=-2*(-1)+a-b+7b=2+4+7b=6+7b=6+7（-2

(2a+5b-3ab)-(a+6b-ab)-(2ab+2b-2a)=2a+5b-3ab-a-6b+ab-2ab-2b+2a=3a-3b-4ab=3(a-b)-4ab=12-12=0

原式=2a+3b-2ab-a-4b-ab-3ab-2b+2a=3a-3b-6ab=3(a-b)-6ab=3*3-6*1=3