已知x²+y²+z²-2x+4y-6z+14=0,则x+y+z的值
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![已知x²+y²+z²-2x+4y-6z+14=0,则x+y+z的值](/uploads/image/z/5261498-26-8.jpg?t=%E5%B7%B2%E7%9F%A5x%26%23178%3B%2By%26%23178%3B%2Bz%26%23178%3B-2x%2B4y-6z%2B14%3D0%2C%E5%88%99x%2By%2Bz%E7%9A%84%E5%80%BC)
已知x²+y²+z²-2x+4y-6z+14=0,则x+y+z的值
已知x²+y²+z²-2x+4y-6z+14=0,则x+y+z的值
已知x²+y²+z²-2x+4y-6z+14=0,则x+y+z的值
(x-1)²+(y+2)²+(z-3)²=0
故x=1,y=-2,z=3
x+y+z=2
由已知得 (x-1)^2+(y+2)^2+(z-3)^2=0
所以: x=1,y=2,z=3
x+y+z=6