vì ( 2x -1)2008>= 0 ( y-2/5)2008 >= 0 ( vì 2008 chẵn)
/ x +y-z/ >=0
=> (2x-1)2008+(y-2/5)2008 +/x+y-z/ >= 0
dấu = xảy ra <=> đồng thời (2x-1)=0, (y-2/5) = 0 , /x+y-z/=0
<=> x=1/2 , y= 2/5 và z = -9/10
tìm x,y,z thuộc N,biết :
a)A=(3x-5)^2006+(y^2-1)^2008+(x-z)^2100=0
b)B=(2x-1)^2008+(y-2:5)^2008+/x+y-z/=0
tìm x,y,z biết: (2x-1)^2008+(y-2/5)^2008+| x+y+z|=0
Tìm x, y, z biết: (2x-1)^2008+(y-2/5)^2008+|x+y-z|=0
(2x-1)^2008+(y-2/5)^2008+ |x+y-z|=0
Tìm x; y ; z biết : (2x-1)^2008 + (y-2/5)^2008 + I x+y-z I = o
tìm xyz
(2x-1)^2008+(y-2/5)^2008+\x+y+z\=0
Tìm x; y; z :
a) \(2009-\left|x-2009\right|=x\)
b) \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
(2x-1)2008 + (y-\(\frac{2}{5}\))\(^{2008}\)+|x+y+z|=0. Tính x,y,z
Tìm x y z biết
\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
