解：原方程變形為?$\frac {x-a-b-c}h8xf99z8w-1+\frac {x-a-b-d}{c}-1+ \frac {x-a-c-d}-1+\frac {x-b-c-d}{a}-1=0$?
∴?$\frac {x-a-b-c-d}h8xf99z8w+\frac {x-a-b-c-d}{c}+\frac {x-a-b-c-d}+\frac {x-a-b-c-d}{a}=0$?
∴?$(x-a-b-c-d)(\frac {1}{a}+\frac {1}+\frac {1}{c}+\frac {1}h8xf99z8w)=0$?
∵?$a$?，?$b$?，?$c$?，?$d$?是正數(shù)
∴?$\frac {1}{a}+\frac {1}+\frac {1}{c}+\frac {1}h8xf99z8w≠0$?
∴?$x-a-b-c-d=0$?
∴?$x=a+b+c+d$?
