设实数x,y满足x^2+2xy-1=0,则x^2+y^2的最小值是？
展开全部
2xy = 1 - x^2,
y = (1-x^2)/(2x),
x^2 + y^2 = x^2 + (1-x^2)^2/(4x^2) = x^2 + (1-2x^2 + x^4)/(4x^2)
= x^2 + x^2/4 + 1/(4x^2) - 1/2
= 5x^2/4 + 1/(4x^2) - 1/2
>= 2[(5/4)(1/4)]^(1/2) - 1/2
= 2*5^(1/2)/4 - 1/2
= [5^(1/2) -1]/2
x^2 + y^2的最小值为[5^(1/2) - 1]/2.
y = (1-x^2)/(2x),
x^2 + y^2 = x^2 + (1-x^2)^2/(4x^2) = x^2 + (1-2x^2 + x^4)/(4x^2)
= x^2 + x^2/4 + 1/(4x^2) - 1/2
= 5x^2/4 + 1/(4x^2) - 1/2
>= 2[(5/4)(1/4)]^(1/2) - 1/2
= 2*5^(1/2)/4 - 1/2
= [5^(1/2) -1]/2
x^2 + y^2的最小值为[5^(1/2) - 1]/2.
已赞过
已踩过<
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
