已知实数x,y,满足x-√x-1=√y+3-y,则x+y最大值是?

题目

已知实数x,y,满足x-√x-1=√y+3-y,则x+y最大值是?

答案

∵x-√x-1=√y+3-y ==>x-√x+1/4+y-√y+1/4=9/2
==>(√x-1/2)²+(√y-1/2)²=9/2
∴设√x-1/2=3cosa/√2,则√y-1/2=3sina/√2
∴x+y=(1/2+3cosa/√2)²+(1/2+3sina/√2)²
=5+3(sina/√2+cosa/√2)
=5+3(sina*cos(π/4)+cosa*sin(π/4))
=5+3*sin(a+π/4)
∵│sin(a+π/4)│≤1
∴│x+y-5│=3│sin(a+π/4)│≤3
==>-3≤x+y-5≤3
==>2≤x+y≤8
故x+y最大值是8.