初三数学-2

设x、y为实数，且x2+xy+y2=1，求x2-xy+y2的取值范围。解 设k=(x^2-xy+y^2)/(x^2+xy+y^2),则 (k-1)x^2+(k+1)xy+(k-1)y^2=0求x2-xy+y2的取值范围,就是求k的取值范围.即 (k+1)^2>=4(k-1)^2故得 3>=k>=1/3