若x²+y²=2,且|x|≠|y|,求1/(x+y)²+1/(x-y)²的最小值
若x²+y²=2,且|x|≠|y|,求1/(x+y)²+1/(x-y)²的最小值
日期：2021-06-07 11:45:13 人气：1
x²+y²=2
令x=√2cost,y=√2sint
1/(x+y)²+1/(x-y)²
=1/(√2cost+√2sint)^2+1/(√2cost-√2sint)^2
=1/[2sin(π/4+t)]^2+1/[2sin(π/4-t)]^2
=1/[2sin(π/4+t)]^2+1/[2cos(π/4+t)]^2
令x=√2cost,y=√2sint
1/(x+y)²+1/(x-y)²
=1/(√2cost+√2sint)^2+1/(√2cost-√2sint)^2
=1/[2sin(π/4+t)]^2+1/[2sin(π/4-t)]^2
=1/[2sin(π/4+t)]^2+1/[2cos(π/4+t)]^2