已知实数x,y满足x2+3x+y-3=0,则x+y的最大值为_____
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![已知实数x,y满足x2+3x+y-3=0,则x+y的最大值为_____](/uploads/image/z/1031335-7-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0x%2Cy%E6%BB%A1%E8%B6%B3x2%2B3x%2By-3%3D0%2C%E5%88%99x%2By%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA_____)
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_____
由题目x²+3x+y-3=0可得
x+y+x²+2x-3=0
x+y=-(x+1)²+4
当x=-1时,x+y取最大值为4
令x+y=a
则y=a-x
所以x²+3x+a-x-3=0
x²+2x+a-3=0
x是实数则△>=0
4-4a+12>=0
a<=8
所以x+y最大值是 8
y=3-x^2-3x
x+y=x+3-x^2-3x
=-x^2-2x+3
=-(x^2+2x-3)
=-((x+1)^2-1-3)
=-((x+1)^2-4)
=-(x+1)^2+4
x=-1,fmax=4
y=-x2-3x+3
x+y=x-x2-3x+3
=-x2-2x+3≥(-12-4)/(-4)=4
上面的做法是对的,但是算错了……
满足x2+3x+y-3=0 确定这些没打错