Question

设x,y是实数,且x^2+xy+y^2=1,求x^2-xy+y^2的取值范围 用换元法计算!

Answer 1

设x^2-xy+y^2=p

∵x^2+xy+y^2=1

∴可知x^2+y^2=(1+p)/2

2xy=(1-p)

∵x^2+y^2≥2|xy|

∴(1+p)/2≥│(1-p)│

∴1/3≤p≤3

∴x^2-xy+y^2∈[1/3,3]