麻烦你们啦,再次深深鞠一躬!已知实数x,y,z满足x+y=5,z^2=xy+y-9,那么x+2y+3z=?我很着急

由条件x=5-y,
因为z^2=xy+y-9=(5-y)y+y-9=6y-y^2-9=-(y-3)^2,
又因为z^2>=0,
所以y=3,x=2，
z^2=2*3+3-9=0，
故x+2y+3z=2+2*3+3*0=8

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
x+2y+3z
=2+6+0
=8