已知实数x,y满足x²+3x+y-3=0,则x+y的最大值为?

已知实数x,y满足x²+3x+y-3=0,则x+y的最大值为?
已知实数x,y满足x²+3x+y-3=0,则x+y的最大值为?

已知实数x,y满足x²+3x+y-3=0,则x+y的最大值为?
由x²+3x+y-3=0得y = 3-x²-3x,
所以 x+y = 3-x² - 2x = -(x+1)² + 4
故最大值为4

∵x² +3x+y-3=0
∴y=-x²-3x+3
∴x+y=-x²-2x+3
=-(x²+2x+1)+3+1
=-(x+1)²+4≤4
∴x+y的最大值是4