解分式方程:1/X+1-1/X+4=1/X+2-1/X+3
2个回答

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

首相,分母不为零:X+1≠0,X+4≠0,X+2≠0,X+3≠0

即x≠-1,-2,-3,-4

两边分别通分:

(x+4-x-1) / [(X+1)(X+4)] = (x+3-x-2) / [(X+2)(X+3) ]

3 / [X^2+5x+4] = 1 / [X^2+5X+6 ]

X^2+5x+4 = 3(X^2+5X)+18

2(X^2+5X)+14=0

X^2+5X+7=0

判别式=5^2-4*7=-3<0,无实数解

在复数范围内:

x=-5/2 ± 根号3/2 i