Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=4k\end{cases}}\)

Khi đó x² + 3y² - 2z² = -729

<=> (2k)² + 3(3k)² - 2(4k)² = -729

=> 4k² + 27k² - 32k² = -729

=> -k² = -729

=> k² = 729

=> k = \(\pm27\)

Khi k = 27 => x = 54 ; y = 81 ; z = 108

Khi k = -27 => x = -54 ; y = -81 ; z = -108

Vậy các cặp (x;y;z) thỏa mãn : (54 ; 81 ; 108) ; (-54;-81;-108)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=4k\end{cases}}\)

Khi đó x² + 3y² - 2z² = -729

<=> ( 2k )² + 3.( 3k )² - 2.( 4k )² = -729

<=> 4k² + 27k² - 32k² = -729

<=> -k² = -729

<=> k² = 729

<=> k = ±27

Với k = 27 => x = 54 ; y = 81 ; z = 108

Với k = -27 => x = -54 ; y = -81 ; z = -108

Ta có :\(\frac{1}{x}=\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

=> \(\frac{1}{x}=\frac{y+z}{2yz}\)

=> 2yz = x(y + z)

=> 2yz - xy - xz = 0

=> (yz - xy) + (yz - xz) = 0

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

=> y(z - x) = -z(y - x)

=> -y(x - z) = -z(y - x) 

=> \(\frac{-z}{-y}=\frac{x-z}{y-x}\Leftrightarrow\frac{z}{y}=\frac{x-z}{y-x}\) 

Đặt \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}=k\Rightarrow\hept{\begin{cases}x=2k\\2y=3k\\3z=4k\end{cases}}\Rightarrow\hept{\begin{cases}x=2k\\y=\frac{3}{2}k\\z=\frac{4}{3}k\end{cases}}\)

Khi đó xyz = -108

<=> \(2k.\frac{3}{2}k.\frac{4}{3}k=-108\)

=>  4k³ = -108

=> k³ = -27

=> k³ = (-3)³

=> k = -3

=> x = -6 ; y = -4,5 ; z = -4

Vậy x = -6 ; y = -4,5 ; z = -4 là giá trị cần tìm