Question

若方程x+2分之x+1-x-1分之x=x^2+x-2分之a的解为正数，求a取值范围

Answer 1

两边乘(x+2)(x-1)
x^2-1-x^2-2x=a
x=(-1-a)/2>0
-1-a>0
a<-1

分母不等于0
所以x≠1
(-1-a)/2≠1
a≠-3

所以a<-1且a≠-3

Answer 2

原式=[y(y+1)(y-1)/(xy+1+x+y)(xy+1-x-y)]*[(x+1)/u(x+y)]
=[y(y+1)(y-1)/(x+1)(y+1)(x-1)(y-1)]*[(x+1)/y(x+y)]
=[y/(x+1)(x-1)]*[(x+1)/y(x+y)]
=1/(x-1)(x+y)
=1/131