Question

实数x、y满足x^2;+xy+y^2;=2,记u=x^;-xy+y^2;,则u的取值范围是yiz

Answer 1

x^2+xy+y^2=2

(x-y)^2+3xy=2

(x-y)^2=2-3xy≥0

xy≤2/3

x^2+xy+y^2=2

(x+y)^2-xy=2

(x-y)^2=2+xy≥0

xy≥-2

所以 -2≤xy≤2/3

x^2-xy+y^2=u

x^2+xy+y^2=2

等式两边相减

-2xy=u-2

因为 -2≤xy≤2/3

所以 4≥u-2≥-4/3

即 2/3≤u≤6