解：原方程变形为$\frac {x-a-b-c}{d}-1+\frac {x-a-b-d}{c}-1+ \frac {x-a-c-d}{b}-1+\frac {x-b-c-d}{a}-1=0$
∴$\frac {x-a-b-c-d}{d}+\frac {x-a-b-c-d}{c}+\frac {x-a-b-c-d}{b}+\frac {x-a-b-c-d}{a}=0$
∴$(x-a-b-c-d)(\frac {1}{a}+\frac {1}{b}+\frac {1}{c}+\frac {1}{d})=0$
∵$a$，$b$，$c$，$d$是正数
∴$\frac {1}{a}+\frac {1}{b}+\frac {1}{c}+\frac {1}{d}≠0$
∴$x-a-b-c-d=0$
∴$x=a+b+c+d$
