文[1]给出了一类带条件的分式型最值问题的一种解法———代“1”法,本文给出这类问题的另一种解法———加零法.例1已知x,y0且x+y=1,求u=1x+16y的最小值.解u=1x+1y6+λ(x+y-1),其中λ为待定的正常数.则u=(1x+λx)+(1y6+λy)-λ≥21x.λx+21y6.λy-λ=10λ-λ,等号成立的充要? [1] gives a solution to a class of conditional fractional-valued problem--generation “1” method. This paper gives an alternative solution to this kind of problem--plus zero method. 1 Known x,y0 and x+y=1, find the minimum value of u=1x+16y. Solution u=1x+1y6+λ(x+y-1), where λ is the normal number to be determined. Then u= (1x+λx)+(1y6+λy)-λ≥21x.λx+21y6.λy-λ=10λ-λ, what does the equal sign hold?