已知x+y=2009,y+2z=2010,z+2x=2011,则x+y+z=多少
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已知x+y=2009,y+2z=2010,z+2x=2011,则x+y+z=多少已知x+y=2009,y+2z=2010,z+2x=2011,则x+y+z=多少已知x+y=2009,y+2z=2010
已知x+y=2009,y+2z=2010,z+2x=2011,则x+y+z=多少
已知x+y=2009,y+2z=2010,z+2x=2011,则x+y+z=多少
已知x+y=2009,y+2z=2010,z+2x=2011,则x+y+z=多少
x+2y=2009…①
y+2z=2010…②
z+2x=2011…③
①+②+③得:
3(x+y+z)=3X2010
x+y+z=2010
x+y=2009
y+2z=2010
z+2x=2011
y=2009-x
z=2011-2x
x+y+z=x+2009-x+2011-2x=2020-2x=2020-(2011-z)=9+z
x+y=9
x+y=2009
有问题啊!!!
x+2y=2009…①
y+2z=2010…②
z+2x=2011…③
①+②+③得:
3(x+y+z)=3X2010
x+y+z=2010
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