Question

实数x，y满足x^2+y^2=1，则2xy/（x+y-1）的取值范围

需详细步骤，谢了

Answer 1

实数x，y满足x²+y²=1，则2xy/（x+y-1）的取值范围
解：∵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

Answer 2

因为2xy=x^2+2xy+y^2-(x^2+y^2)=(x+y)^2-1，设z=x+y
则2xy/(x+y-1)=[(x+y)^2-1]/(x+y-1)=(z^2-1)/(z-1)=(z+1)(z-1)/(z-1)=z+1
所以，只需求z=x+y的范围即可。首先x+y≠1，即z≠1，z+1≠2
然后，z^2=(x+y)^2=x^2+2xy+y^2=-(x^2-2xy+y^2)+2=-(x-y)^2+2
所以z^2<=2,-√2<=z<=√2，1-√2<=z+1<=1+√2
∴所求式子的取值范围为[1-√2,2)∪(2,1+√2]，
或者说1-√2<=所求式<=1+√2，且所求式≠2