Question

三道题，

Answer 1

2.∵x+y+z=5，∴(x+y+z)2=25，即x2+y2+z2+2(xy+yz+zx)=25
又∵xy+yz+zx=3，∴x2+y2+z2=19，
(5-y-z)2+y2+z2=19
y2+z2-5y-5z+yz+3=0
y2+(z-5)y+z2-5z+3=0
∵y是实数，∴△=(z-5)2-4(z2-5z+3)≥0
即3z2-10z-13≤0
∴-1≤z≤13/3
所以最大值为13/3

追问

3题4题呢

追答

4.2x2-6x+y2=0
x2+y2=6x-x2
则原式=6x-x2+2x
=-x2+8x
=-(x2-8x+16）+16
=-(x-4)2+16
当x=4时，原式max=16

Answer 2

2. x+y+z=5 (1)
xy+yz+zx=3 (2)

3=zx+(x+z)(5-x-z)
x^2+(z-5)x+z^2+5z+3=0
(z-5)^2-4(z^2+5z+3)>=0
-3z^2-30z+13>=0
3z^2+30z-13<=0
(-30-√1056)/6<=z<=(-30+√1056)/6

z最大值是(-30+√1056)/6=(-15+2√66)/3

Answer 3

jss*8

Answer 4