Giải thích các bước giải:
Ta có:
$\begin{array}{l}
y = \sqrt {{x^2} - 2} \\
\Rightarrow y' = \dfrac{x}{{\sqrt {{x^2} - 2} }}\\
\Rightarrow y'' = \dfrac{{\sqrt {{x^2} - 2} - x.\dfrac{x}{{\sqrt {{x^2} - 2} }}}}{{{{\left( {\sqrt {{x^2} - 2} } \right)}^2}}} = \dfrac{{ - 2}}{{{{\left( {\sqrt {{x^2} - 2} } \right)}^3}}}
\end{array}$
Khi đó:
$\begin{array}{l}
y''{y^3} + {\left( {y'} \right)^2}\\
= \dfrac{{ - 2}}{{{{\left( {\sqrt {{x^2} - 2} } \right)}^3}}}.{\left( {\sqrt {{x^2} - 2} } \right)^3} + {\left( {\dfrac{x}{{\sqrt {{x^2} - 2} }}} \right)^2}\\
= - 2 + \dfrac{{{x^2}}}{{{x^2} - 2}}\\
= \dfrac{{{x^2}}}{{{x^2} - 2}} - 1 - 1\\
= \dfrac{2}{{{x^2} - 2}} - 1\\
= \dfrac{2}{{{{\left( {\sqrt {{x^2} - 2} } \right)}^2}}} - 1\\
= \dfrac{2}{{{y^2}}} - 1
\end{array}$
