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已知x^2+y^2+z^2=1,x+2y+3z=14根号,x+y+z=?
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已知x^2+y^2+z^2=1,x+2y+3z=14根号,x+y+z=?
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答案和解析
(x+2y+3z)^2=14
14(x^2+y^2+z^2)=(x+2y+3z)^2,化简得（2x-y)^2+(3x-z)^2+(2z-3y)^2=0
所以y=2x,z=3x,代入得x+y+z=3*14*(1/2)/7
14(x^2+y^2+z^2)=(x+2y+3z)^2,化简得（2x-y)^2+(3x-z)^2+(2z-3y)^2=0
所以y=2x,z=3x,代入得x+y+z=3*14*(1/2)/7
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