一道数学题，高手进~~！谢谢了！！

解：
已知tan(x+y)=m/(2+m),tanx*tany=m-1,且tan²(x-y)=9则
tan(x-y)=±√9=±3

tan(x-y)
=（tanx-tany)/（1+tanx*tany)
=（tanx-tany)/[1+(m-1)]
=（tanx-tany)/m
=±3
tanx-tany=±3m
讨论：(A)tanx-tany=3m......(1)
tan(x+y)
=（tanx+tany)/（1-tanx*tany)
=（tanx+tany)/]1-(m-1)]
=（tanx+tany)/(2-m)
=m/(2+m)

tanx+tany=m*(2-m)/(2+m)......(2)
(1)+(2):tanx=m*(m+4)/(2+m)
(2)-(1):tany=-2m*(1+m)/(2+m)

tanx*tany=m-1
[m*(m+4)/(2+m)]*[-2m*(1+m)/(2+m)]=m-1
m^2*(2m^2+11m+11)-4=0
不懂解这一方程，但可知-5<m<-4
（B）tanx-tany=-3m......(3)结果的tanx、tany与（A）互调，即求m的结果相同。