\(5x=4y=2z\)

\(\Leftrightarrow\frac{5x}{20}=\frac{4y}{20}=\frac{2z}{20}\)

\(\Leftrightarrow\frac{x}{4}=\frac{y}{5}=\frac{z}{10}\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\frac{x}{4}=\frac{y}{5}=\frac{z}{10}=\frac{x-y+z}{4-5+10}=\frac{-18}{8}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{4}=-2\\\frac{y}{5}=-2\\\frac{z}{10}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=-10\\z=-20\end{matrix}\right.\)

Lại có :

\(A=\left(\frac{2}{x}+\frac{5}{y}+\frac{5}{z}\right)^{2016}\)

\(=\left(\frac{2}{-8}+\frac{5}{-10}+\frac{5}{-20}\right)^{2016}\)

\(=1\)

Vậy....

Đặt 5x=4y=2z=k suy ra \(x=\frac{k}{5};y=\frac{k}{4};z=\frac{k}{2}\)

Ta có : 

x-y+z=-18

\(\frac{k}{5}-\frac{k}{4}+\frac{k}{2}=-18\)

\(k.\left(\frac{1}{5}-\frac{1}{4}+\frac{1}{2}\right)=-18\)

\(k.\frac{9}{20}=-18\)

k = -40 suy ra x = -8 ; y = -10 ; z = -20

Ta có:

\(A=\left(\frac{2}{x}+\frac{5}{y}+\frac{5}{z}\right)^{2016}=\left(\frac{2}{-8}+\frac{5}{-40}+\frac{5}{-20}\right)^{2016}=\left(\frac{5}{-8}\right)^{2016}=0\)

Ta có: x^2+2y^2+z^2-2xy-2y-4z+5=0

<=> ( x^2 - 2xy + y^2 ) + ( y^2 - 2y +1 ) + ( z^2 - 4z + 4 ) = 0

<=> ( x - y )^2 + ( y - 1 )^2 + ( z - 2 )^2 = 0

=> x - y = 0 và y - 1 = 0 và z - 2 = 0

<=> x = y = 1 và z = 4

Nên P = 1

ta có x^2 + 2y^2 +z^2 -2xy -2y -4z +5 =0

=> (x^2 - 2xy +y^2) + (y^2 -2y +1) + (z^2 -4z +4) =0

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

=> x=y , y=1 , z=2 

=> A= (1-1)^2018 + (1-1)^2019 + ( 2-1)^2020 => A= 1

nghĩ thế !