已知m∈R,α,β∈(-π/2,π/2),且tanα,tanβ是方程x^2-mx+1=m的两个根,求α+β

由韦达定理,可得：
tanα+tanβ=m
tanαtanβ=1-m
方程有两个实数根,则△≥0
△=(-m)²-4×1×(1-m)
=m²+4m-4
m²-4m-4≥0
(m²-4m+4)-8≥0
(m-2)²-(2√2)²≥0
[(m-2)+2√2][(m-2)-2√2]≥0
(m+2√2-2)(m-2√2-2)≥0
解得 m∈(-∞,-2√2+2]∪[2√2+2,+∞)
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)
=m/[1-(1-m)]
=m/m
=1
∵α,β(-π/2,π/2)
∴α+β∈(-π,π)
∴α+β=-3π/4,或α+β=π/4