Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{-1}{z}\)
\(\Leftrightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+3\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}+3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y^2}+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\dfrac{-1}{z}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
\(\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3}{xyz}\cdot xyz\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{zy}{z^2}=3\)
hay B=3
Ta có: $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$
$⇒\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}$
$⇔(\dfrac{1}{x}+\dfrac{1}{y})^3=-\dfrac{1}{z^3}$
$⇔\dfrac{1}{x^3}+\dfrac{1}{y^3}+3\dfrac{1}{x}.\dfrac{1}{y}(\dfrac{1}{x}+\dfrac{1}{y})=-\dfrac{1}{z^3}$
$⇔\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}.\dfrac{1}{y}(\dfrac{1}{x}+\dfrac{1}{y})$
$⇔\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=3.\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}$
do $\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}$
Có: $B=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}$
$=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}$
$=xyz.(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})$
$=xyz.3.\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}$
$=3$
Vậy $B=3$ với $x;y;z$ t/m đề
