根号下a+2√a-1 + 根号下a-2√a-1(a≥1)
=√[a+ 2√(a-1)]+√[a-2√(a-1)]
=√[(a-1)+ 2√(a-1)+ 1]+√[(a-1)- 2√(a-1) +1]
=√[√(a-1)+ 1]^2+√[√(a-1)- 1]^2,因为a≥1,
=√(a-1)+ 1 +√(a-1)-1
=2√(a-1)
根号下a+2√a-1 + 根号下a-2√a-1(a≥1)
=√[a+ 2√(a-1)]+√[a-2√(a-1)]
=√[(a-1)+ 2√(a-1)+ 1]+√[(a-1)- 2√(a-1) +1]
=√[√(a-1)+ 1]^2+√[√(a-1)- 1]^2,因为a≥1,
=√(a-1)+ 1 +√(a-1)-1
=2√(a-1)