Đáp án:
$S=3$
Giải thích các bước giải:
ĐKXĐ: $x,y,z\ne 0$
Ta có:
$\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{z} = 0 \Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{z} = \dfrac{1}{x} + \dfrac{1}{y}\\
\dfrac{1}{x} = \dfrac{1}{z} - \dfrac{1}{y}\\
\dfrac{1}{y} = \dfrac{1}{z} - \dfrac{1}{x}
\end{array} \right.$
Khi đó:
$\begin{array}{l}
S = \dfrac{{xy}}{{{z^2}}} - \dfrac{{yz}}{{{x^2}}} - \dfrac{{zx}}{{{y^2}}}\\
= xy.{\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right)^2} - yz.{\left( {\dfrac{1}{z} - \dfrac{1}{y}} \right)^2} - zx.{\left( {\dfrac{1}{z} - \dfrac{1}{x}} \right)^2}\\
= xy\left( {\dfrac{1}{{{x^2}}} + \dfrac{2}{{xy}} + \dfrac{1}{{{y^2}}}} \right) - yz\left( {\dfrac{1}{{{z^2}}} - \dfrac{2}{{yz}} + \dfrac{1}{{{y^2}}}} \right) - zx\left( {\dfrac{1}{{{z^2}}} - \dfrac{2}{{xz}} + \dfrac{1}{{{x^2}}}} \right)\\
= \dfrac{y}{x} + 2 + \dfrac{x}{y} - \dfrac{y}{z} + 2 - \dfrac{z}{y} - \dfrac{x}{z} + 2 - \dfrac{z}{x}\\
= 6 + \left( {\dfrac{y}{x} - \dfrac{y}{z}} \right) + \left( {\dfrac{x}{y} - \dfrac{x}{z}} \right) - \left( {\dfrac{z}{y} + \dfrac{z}{x}} \right)\\
= 6 + y\left( {\dfrac{1}{x} - \dfrac{1}{z}} \right) + x\left( {\dfrac{1}{y} - \dfrac{1}{z}} \right) - z\left( {\dfrac{1}{y} + \dfrac{1}{x}} \right)\\
= 6 + y.\left( {\dfrac{{ - 1}}{y}} \right) + x.\left( {\dfrac{{ - 1}}{x}} \right) - z.\dfrac{1}{z}\\
= 6 - 1 - 1 - 1\\
= 3
\end{array}$
Vậy $S=3$
