已知x,y属于R,且x²+y²≤1,求|x+y|的取值范围
∵(x-y)²=x²-2xy+y²≥0,∴2xy≤x²+y²≤1,
于是(x+y)²=x²+y²+2xy≤1+1=2
故0≤│x+y│≤√2