已知x,y,z满足x+y=5,z^2=xy+y-9 求x+2y+3z的值

x=5-y
z2=(5-y)y+y-9
=6y-y2-9
=-(9-6y+y2)
=-(y-3)2
由题意，只有当该项为0时等式成立
得y=3
那么z=0 x=2
即原式=2+6+0=8

x=5-y z2=-(y-3)2 y=3 z=0 x=2

=2+6+0=8

z^2=(5-y)y + y -9 = 5y-y^2+y-9 = -(y^2-6y+9) = -(y-3)^2

所以，z^2 = -(y-3)^2, 所以z=0, y-3=0

所以 y=3, x=2

所以 x+2y+3z = 2 + 2*3 + 0*3 = 8