ta có:

(2x-4)^2>0

|y-5|>0

(x+y-z)^6>0

=>(2x-4)^2+|y-5|+(x+y-z)^6>0

mà theo đề:(2x-4)^2+|y-5|+(x+y-z)^6=0

=>(2x-4)^2=|y-5|=(x+y-z)^6=0

+)(2x-4)^2=0=>2x-4=0=>2x=4=>x=2

+)|y-5|=0=>y-5=0=>y=5

+)(x+y-z)^6=0=>x+y-z=0

thay x=2;y=5 vào x+y-z=0

=>2+5-z=0

=>7-z=0

=>z=7

vậy..

nhớ tick

(2x-4)²+|y-5| + (x+y+z)⁶ = 0 

Ta thay: (2x-4)²  ;   |y+5|     ;   ( x+y+z )⁶  lon hon hoac bang 0 ( vi so mu chan va gia tri tuyet doi )

Ma: (2x-4)² + |y-5| + (x+y+z)⁶ = 0 

=> (2x-4)² = |y-5| = (x+y+z)⁶  ( = 0 )

Vay: z can tim la -7

Ta thấy:\(\begin{cases}\left(2x-4\right)^2\\\left|y-5\right|\\\left(x+y-z\right)^6\end{cases}\ge0\)

\(\Rightarrow\left(2x-4\right)^2+\left|y-5\right|+\left(x+y-z\right)^6\ge0\)

Dấu = khi

<=>2x=4 =>x=2

=>y-5=0 =>y=5

Ta thay x,y vào (x+y-z)⁶=0 đc:

(2+5-z)⁶=0 =>7-z=0 =>z=7

Vậy giá trị của z=7