Đáp án:
$\begin{array}{l}
2)y = \frac{{x - 3}}{{x + 4}}\\
\Rightarrow y' = \frac{7}{{{{\left( {x + 4} \right)}^2}}}\\
\Rightarrow y'' = 7.\left( { - 2} \right).\frac{1}{{{{\left( {x + 4} \right)}^3}}} = \frac{{ - 14}}{{{{\left( {x + 4} \right)}^3}}}\\
Do:2y{'^2} = \frac{{14}}{{{{\left( {x + 4} \right)}^4}}}\\
\left( {y - 1} \right).y'' = \left( {\frac{{x - 3}}{{x + 4}} - 1} \right).\frac{{ - 14}}{{{{\left( {x + 4} \right)}^3}}} = \frac{{14}}{{{{\left( {x + 4} \right)}^4}}}\\
\Rightarrow 2y{'^2} = \left( {y - 1} \right)y''\\
3)\\
y = \sqrt {x + 1} = {\left( {x + 1} \right)^{\frac{1}{2}}}\\
\Rightarrow y' = \frac{1}{2}.{\left( {x + 1} \right)^{ - \frac{1}{2}}}\\
\Rightarrow y'' = \frac{1}{2}.\left( { - \frac{1}{2}} \right).{\left( {x + 1} \right)^{\frac{{ - 3}}{2}}} = \frac{{ - 1}}{4}.{\left( {x + 1} \right)^{ - \frac{3}{2}}}\\
\Rightarrow 4{y^3}.y'' = 4.{\left( {x + 1} \right)^{\frac{3}{2}}}.\left( { - \frac{1}{4}} \right).{\left( {x + 1} \right)^{ - \frac{3}{2}}} = - 1\\
\Rightarrow 4{y^3}.y'' + 1 = - 1 + 1 = 0
\end{array}$
