實(shí)數(shù)x,y滿足x²+y²=1,則2xy/（x+y-1）的取值范圍
∵x²+y²=(x+y)²-2xy=1,故2xy=(x+y)²-1=(x+y+1)(x+y-1),
又∵x²+y²=1,∴可令x=cost,y=sint；
∴2xy/(x+y-1)=(x+y+1)(x+y-1)/(x+y-1)=x+y+1=cost+sint+1=(√2)sin(t+π/4)+1
而-√2≦(√2)sin(t+π/4)≦√2
1-√2≦2xy/(x+y-1)≦1+√2