Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\left| {7x - 5y} \right| \ge 0,\,\,\,\,\forall x,y\\
\left| {2z - 3x} \right| \ge 0,\,\,\,\forall x,z\\
\left| {xy + yz + zx - 4500} \right| \ge 0,\,\,\,\forall x,y,z\\
 \Rightarrow A = \left| {7x - 5y} \right| + \left| {2z - 3x} \right| + \left| {xy + yz + zx - 4500} \right| \ge 0,\,\,\,\forall x,y,z\\
 \Rightarrow {A_{\min }} = 0 \Leftrightarrow \left\{ \begin{array}{l}
7x - 5y = 0\\
2z - 3x = 0\\
xy + yz + zx = 4500
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
7x = 5y\\
2z = 3x\\
xy + yz + zx = 4500
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = \frac{7}{5}x\\
z = \frac{3}{2}x\\
xy + yz + zx = 4500
\end{array} \right.\\
xy + yz + zx = 4500\\
 \Leftrightarrow x.\frac{7}{5}x + \frac{7}{5}x.\frac{3}{2}x + \frac{3}{2}x.x = 4500\\
 \Leftrightarrow 5{x^2} = 4500\\
 \Leftrightarrow {x^2} = 900\\
 \Leftrightarrow x =  \pm 30 \Rightarrow y =  \pm 42;\,\,\,z =  \pm 45
\end{array}\)

Ta có $ |7x -5y| \ge 0;   |2z -3x| \ge 0 ;  |xy +yz+ zx - 4500| \ge 0 ∀ x ; y; z $

$ \to A \ge 0 $

Dấu $=$ xảy ra khi $ 7x - 5y = 0 ; 2z-3x = 0 ; xy +yz + xz - 4500 = 0$

$ 7x -5y = 0 \to 7x = 5y \to \dfrac{x}{5} = \dfrac{y}{7}$

$ 2z -3x=0 \to 3x = 2z \to \dfrac{x}{2} = \dfrac{z}{3}$

$ \to \dfrac{x}{10} = \dfrac{y}{14} = \dfrac{z}{15}$

$ xy + yz +xz = 4500$

$ \to ( \dfrac{x}{10})^2 = ( \dfrac{y}{14})^2 = ( \dfrac{z}{15})^2 = \dfrac{xy}{140} = \dfrac{yz}{210} = \dfrac{xz}{150}$

$ = \dfrac{xy+yz+xz}{140+210+150} = \dfrac{4500}{500} = 9$

$ \to $

$\dfrac{x^2}{100} = 9 \to x = ± 30$

$ \dfrac{y^2}{196} = 9 \to y =± 42$

$ \dfrac{z^2}{225} = 9 \to z = ± 45$