Question

八上数学分式题：先化简再求值a-b分之a+b-a+b分之a-b-a²-b²分之3ab，其中a=3，b=-2

计算x²-1分之a+b+x+1分之b=
x-1分之2-2-2x分之1-3x=
已知x²+xy+y²=0，则x分之y+y分之x=
计算x²-y²分之x+y+y²-x²分之2y

a²-2a+1分之a²-1+a²-1分之a-a²

要过程、结果！谢谢大家。

Answer 1

(a+b)/(a-b) -(a-b)/(a+b) -3ab/(a^2-b^2)
=(a+b)^2/[(a-b)*(a+b)] -(a-b)^2/[(a+b)*(a-b) ] -3ab/(a^2-b^2)
=[(a+b)^2-(a-b)^2-3ab]/(a^2-b^2)
=ab/(a^2-b^2)
=3*(-2)/[3^2 -(-2)^2]
=-6/(9-4)
=-6/5

(a+b)/(x^2-1) + b/(x+1)
=(a+b)/[(x+1)(x-1)]+b(x-1)/[(x+1)(x-1)]
=(a+b+bx-b)/(x^2-1)
=(a+bx)/(x^2-1)

2/(x-1) -(1-3x)/(2-2x)
=2/(x-1)-(3x-1)/2(x-1)
=4/2(x-1)-(3x-1)/2(x-1)
=(4-3x+1)/2(x-1)
=-3(x-1)/2(x-1)
=-3/2

x^2+xy+y^2=0 两边同时除以xy
x/y+1+y/x=0
x/y +y/x=-1

(x+y)/(x^2-y^2) +2y/(y^2-x^2)
=(x+y)/(x^2-y^2) -2y/(x^2-y^2)
=(x+y-2y)/(x^2-y^2)
=(x-y)/[(x-y)(x+y)]
=1/(x+y)

(a^2-1)/(a^2-2a+1) +(a-a^2)/(a^2-1)
=(a-1)(a+1)/(a-1)^2 +(a-a^2)/(a^2-1)
=(a+1)/(a-1) + (a-a^2)/(a-1)(a+1)
=(a+1)^2/[(a-1)(a+1)] +(a-a^2)/(a-1)(a+1)
=[(a+1)^2 +(a-a^2)]/[(a-1)(a+1)]
=(3a+1)/(a^2-1)

Answer 2

(a+b)/(a-b) -(a-b)/(a+b) -3ab/(a^2-b^2)=(a+b)^2/[(a-b)*(a+b)] -(a-b)^2/[(a+b)*(a-b) ] -3ab/(a^2-b^2)
=[(a+b)^2-(a-b)^2-3ab]/(a^2-b^2)=ab/(a^2-b^2)=3*(-2)/[3^2 -(-2)^2]=-6/(9-4)=-6/5