x-y+z=0(4-x)^2+(2-y)^2=z^2︳x+y-3︳(这是绝对值)/根号2=z

Question

x-y+z=0
(4-x)^2+(2-y)^2=z^2
︳x+y-3︳(这是绝对值)/根号2=z

Answer 1

由3）得：x+y-3=√2z,4）
或x+y-3=-√2z 5）
由4）+1）得：2x+z-3=√2z,得：x=3/2+(√2-1)z/2
故y=x+z=3/2+(√2+1)z/2
将x,y代入2）得：[5/2-(√2-1)z/2]^2+[1/2-(√2+1)z/2]^2=z^2
25+(√2-1)^2z^2-10(√2-1)z+1+(√2+1)^2z^2-2(√2+1)z=4z^2
化简得：z^2-2z(3√2-2)+13=0,此方程判别式小于0,没实根
由5）+1）得：2x+z-3=-√2z,得：x=3/2-(√2+1)z/2
故y=x+z=3/2-(√2-1)z/2
将x,y代入2）得：[5/2+(√2+1)z/2]^2+[1/2+(√2-1)z/2]^2=z^2
25+(√2+1)^2z^2+10(√2+1)z+1+(√2-1)^2z^2+2(√2-1)z=4z^2
化简得：z^2+2z(3√2+2)+13=0
因z为非负数,解之得：z=-(3√2+2)+√(9+12√2)
从而得x=3/2-(√2+1)z/2,y=3/2+(√2+1)z/2
故只有上面1组解.