解:∵在△ABC中,AB=-√2+√6,∠C=30°
设∠A>∠B
过A点作AD⊥BC,交BC于D点
在直角△ACD中
∠C=30°,AD=AC/2,CD=AC*cos30°=(√3/2)*AC
在直角△ABD中
BD^2=AB^2-AD^2
=(-√2+√6)^2-(AC/2)^2
=8-4√3-AC^2/4
BD=√(8-4√3-AC^2/4)
BC=CD+BD=(√3/2)*AC+√(8-4√3-AC^2/4)
AC+BC
=AC+(√3/2)*AC+√(8-4√3-AC^2/4)
=(1+√3/2)*AC+√(8-4√3-AC^2/4)
设AC+BC=s,AC=x,则
s=(1+√3/2)x+√(8-4√3-x^2/4)
s-(1+√3/2)x=√(8-4√3-x^2/4)
[s-(1+√3/2)x]^2=8-4√3-x^2/4
(2+√3)x^2-(2+√3)sx+s^2-4(2-√3)=0
x^2-sx+[s^2-4(2-√3)]/(2+√3)=0
判别式△=(-s)^2-4*[s^2-4(2-√3)]/(2+√3)≥0
s^2≤16
∵s>0
∴(s)max=4
∴(AC+BC)max=4
参考资料:百度一下