求解5道初中数学分式题

Question

1 已知y分之x=3，则y^2分之x^2+xy的值是多少
2 4b^2分之b^2-9a^2除以2b^3分之9ab-3b^2 计算
3 x^2+4x+4分之x^2-6x+9除以x分之x^2-9乘以3-x分之x^2+3x 其中x=2，先分解再求值
4（xy分之x^2-y^2)^2除以（x+y)^2乘以（x-y分之x）^3，其中x=3,y=2,先分解再求值
5 已知x^2-x-2分之3x+4=x-2分之A-x+1分之B，AB为常数，求4A-B的值
求速度，急急急

Answer 1

1 已知y分之x=3，则y^2分之x^2+xy的值是多少
x/y=3
x=3y
(x^2+xy)/y^2
=[(3y)^2+3y*y]/y^2
=(9+3)y^2/y^2
=12
2 4b^2分之b^2-9a^2除以2b^3分之9ab-3b^2 计算
[(b^2-9a^2)/(4b^2)]/[(9ab-3b^2)/(2b^3)]
=[(b+3a)(b-3a)/(4b^2)×(2b^3)/[3b(3a-b)]
=-3b(b+3a)*b/2
=-3/2b^3-9/2ab^2
3 x^2+4x+4分之x^2-6x+9除以x分之x^2-9乘以3-x分之x^2+3x 其中x=2，先分解再求值
[(x^2-6x+9)/(x^2+4x+4)]/[(x^2-9)/x]×[(x^2+3x)/(3-x)]
=[(x-3)^2/(x+2)^2]×x/[(x+3)(x-3)]×[-x(x+3)/(x-3)]
=-x^2/(x+2)^2
=-[x/(x+2)]^2
=-[2/(2+2)]^2
=-1/4
4（xy分之x^2-y^2)^2除以（x+y)^2乘以（x-y分之x）^3，其中x=3,y=2,先分解再求值
[(x^2-y^2)/(xy)]^2/(x+y)^2×[x/(x-y)]^3
=[(x+y)^2(x-y)^2/(xy)^2]/(x+y)^2×[x^3/(x-y)^3]
=x/[y^2(x-y)]
=3/[2^2(3-2)]
=3/4
5 已知x^2-x-2分之3x+4=x-2分之A-x+1分之B，AB为常数，求4A-B的值
(3x+4)/(x^2-x-2)=A/(x-2)-B/(x+1)
(3x+4)/(x^2-x-2)=[A(x+1)-B(x-2)]/(x^2-x-2)
(3x+4)/(x^2-x-2)=[(A-B)x+(A+2B)]/(x^2-x-2)

A-B=3..........(1)
A+2B=4.......(2)
(2)-(1)：3B=1
B=1/3
代入(1)：A-1/3=3
A=10/3
4A-B
=4×10/3-1/3
=(40-1)/3
=13