tanA,tanB是方程mx^2-2√(7m-3) x+2m=0的两个实数根,求：tan(A+B)的最大值

tanA,tanB是方程mx^2-2√(7m-3) x+2m=0的两个实数根,
△=[-2√(7m-3)]²-4m×2m=4(7m-3-2m²)≥0,
2m²-7m+3≤0,(2m-1)(m-3)≤0,1/2≤m≤3,
由根与系数关系得tanA+tanB=2√(7m-3)/m>0,tanAtanB=2m/m=2>0,
则tanA>0,tanB>0,tanA+tanB≥2√(tanAtanB)=2√2,
当tanA=tanB=√2,时tanA+tanB=2√(7m-3)/m=2√2,7m-3=2m²,-2m²+7m-3=0,m=1/2或m=3
即当m=1/2或m=3时,tanA+tanB取最小值2根号2,
tan(A+B)=(tanA+tanB)(1-tanAtanB)=-(tanA+tanB)取最大值-2根号2