Question

biết x;y;z thoả mãn: \(\left(2x-y\right)^2+\left(y-2\right)^2+\sqrt{\left(x+y+z\right)^2}=0\) khi đó x+y-z=

Answer 1

pt<=>\(\left(2x-y\right)^2+\left(y-2\right)^2+\)/x+y+z/=0.

<=> \(\int^{2x-y=0}_{\int^{y-2=0}_{x+y+z=0}}\Leftrightarrow\int^{\int^{y=2}_{x=1}}_{x=-1-2=-3}\)

Answer 2

pt<=> 

2x-y=0(1)y-2=0x+y+z=0

=> x+y+z=0(đpcm)

Answer 3

giai nhầm

pt<=> vì (2x-y)^2;(y-2)^2;căn((x+y+z)^2) >=0

vi pt=0 =>

y-2=0 =>y=22x-y=0  =>x=1x+y+z=0 =>z=-3

vây: x+y-z=2+1-(-3)=6