Question

2x+3y+z=13，4x^2+9y^2+z^2-2x+15y+3z=82，求x,y,z的值

Answer 1

2x+3y+z=13推出(2x-1/2)+(3y+5/2)+(z+3/2)=33/2

4x^2+9y^2+z^2-2x+15y+3z=82推出(2x-1/2)^2+(3y+5/2)^2+(z+3/2)^2=363/4

由柯西不等式得 (
(2x-1/2)+(3y+5/2)+(z+3/2))^2<=((2x-1/2)^2+(3y+5/2)^2+(z+3/2)^2)(1+1+1)

左式等于1089/4
右式等于1089/4
不等式取等号

所以2x-1/2=3y+5/2=z+3/2=11/2即x=3,y=1,z=4

Answer 2

2x+3y+z=13推(2x-1/2)+(3y+5/2)+(z+3/2)=33/2

4x^2+9y^2+z^2-2x+15y+3z=82推出(2x-1/2)^2+(3y+5/2)^2+(z+3/2)^2=363/4

由柯西不等式得 (
(2x-1/2)+(3y+5/2)+(z+3/2))^2<=((2x-1/2)^2+(3y+5/2)^2+(z+3/2)^2)(1+1+1)

左式等于1089/4
右式等于1089/4
不等式取等号

所以2x-1/2=3y+5/2=z+3/2=11/2即x=3,y=1,z=4

Answer 3

求三个实数x,y,z使得它们满足下列方程组:
2x+3y+z=13
(1)
4x^2+9y^2+z^2-2x+15y+3z=82。
(2)
解
(1)+(2)得:
4x^2+9y^2+z^2+18y+4z=95
<==>
(2x)^2+(3y+3)^2+(z+2)^2=108
(3)
由均值不等式可a+b+c=<√[3(a^2+b^2+c^2)]得:
2x+3y+3+z+2=<√(3*108)=18
<==>
2x+3y+z=<13。
(4)
对比(1)与(4)式得到:2x=3y+3=z+2=6。
故x=3,y=1,z=4是方程唯一解。