If the real numbers x, y and z satisfy x+y+z = 5 and xy+yz+zx = 3, the maximum value of z is: 13/3.

Solution:

∫xy+yz+zx = 3，x+y+z=5

∴x+y=5-z

∴2xy=6-2(y+x)z=6-2(5-z)z=2z^2- 10z+6

∴2*(xy+yz+zx)=6

∫x+y+z = 5

∴(x+y+z)^2=25

x^2+y^2+z^2+2*(xy+xz+yz)=25

x^2+y^2+z^2= 19

∵(x-y)^2≥0，

x^2+y^2-2xy≥0，

x^2+y^2≥2xy，

When ∴ x 2+y 2 = 2xy, z 2 has the maximum value.

∴z^2+2xy= 19，

∴z^2+2z^2- 10z+6= 19，

3z^2- 10z- 13=0

z^2- 10z/3- 13/3=0

(z-5/3)^2-(8/3)^2=0

(z- 13/3)*(z+ 1)=0

z 1= 13/3

z2=- 1

z 1 & gt； z2

Therefore, the maximum value of z = 13/3.

A: If the real numbers x, y and z satisfy X+Y+Z = 5 and XY+YZ+ZX = 3, the maximum value of z is 13/3.