若2{√x+√(y-1)+√(z-2)}=x+y+z,求x+y+z的算术平方根.----------------------------------------------------------------

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若2{√x+√(y-1)+√(z-2)}=x+y+z,求x+y+z的算术平方根.------------------------------------------------------------

若2{√x+√(y-1)+√(z-2)}=x+y+z,求x+y+z的算术平方根.----------------------------------------------------------------
若2{√x+√(y-1)+√(z-2)}=x+y+z,求x+y+z的算术平方根.
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若2{√x+√(y-1)+√(z-2)}=x+y+z,求x+y+z的算术平方根.----------------------------------------------------------------
由已知条件,可以移项合并得到
(√x -1)^2 +(√(y-1) -1)^2 +(√(z-2)-1)^2=0
所以
√x -1 =0 => x =1
√(y-1) -1 =0 => y=2
√(z-2)-1 =0 =>z=3
所以x+y+z的算术平方根= √6