y=(x²+2)/√(x²+1)
=(x²+1+1)/√(x²+1)
=√(x²+1)+1/√(x²+1)
≥2√1
=2
当且仅当x²+1=1/(x²+1),即x=0时取等
所以值域为[2,+∞)=√(x²+1)+1/√(x²+1)这一步为什么？y=(x²+1+1)/√(x²+1)=(x²+1)/√(x²+1)+1/√(x²+1)=[√(x²+1)]²/√(x²+1)+1/√(x²+1)=√(x²+1)+1/√(x²+1)