Question

解分式方程:(1)1/(x+6)+1/(x+4)=1/(x+7)+1/(x+3);(2)1/(x-3)+1/(x-7)=1/(x-4)+1/(x-6).

Answer 1

1)设x+5=y
1/(y+1)+1/(y-1)=1/(y+2)+1/(y-2)
2y/(y+1)(y-1)=2y/(y+2)(y-2)
y(y+2)(y-2)=y(y+1)(y-1)

y(y^2-4)=y（y^2-1)
y=0
x+5=0
x=-5
2)设x-5=y
1/(y+2)+1/(y-2)=1/(y+1)+1/(y-1)

2y/(y+2)(y-2)=2y/(y+1)(y-1)

y(y+1)(y-1)=y(y+2)(y-2)

y=0
x-5=0
x=5

追问

不是这样算的吧，(1)应该化成:1/(x+6)+1/(x+7)=1/(x+4)+1/(x+3)
(2)应该化成:1/(x-3)+1/(x-4)=1/(x-7)+1/(x-6)

追答

你可以试试，没我的算法简单呦