Đáp án:
\(\dfrac{{{y^{\left( 3 \right)}}}}{{{y^{\left( 2 \right)}}}} = - \dfrac{6}{{2x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = \dfrac{{2x + 1 - 2\left( {x - 1} \right)}}{{{{\left( {2x + 1} \right)}^2}}} = \dfrac{3}{{{{\left( {2x + 1} \right)}^2}}}\\
y'' = {y^{\left( 2 \right)}} = \dfrac{{ - 2.2.\left( {2x + 1} \right).3}}{{{{\left( {2x + 1} \right)}^4}}}\\
= \dfrac{{ - 12\left( {2x + 1} \right)}}{{{{\left( {2x + 1} \right)}^4}}} = \dfrac{{ - 12}}{{{{\left( {2x + 1} \right)}^3}}}\\
{y^{\left( 3 \right)}} = \dfrac{{ - 3.2.{{\left( {2x + 1} \right)}^2}.\left( { - 12} \right)}}{{{{\left( {2x + 1} \right)}^6}}}\\
= \dfrac{{72}}{{{{\left( {2x + 1} \right)}^4}}}\\
\dfrac{{{y^{\left( 3 \right)}}}}{{{y^{\left( 2 \right)}}}} = \dfrac{{72}}{{{{\left( {2x + 1} \right)}^4}}}:\left[ {\dfrac{{ - 12}}{{{{\left( {2x + 1} \right)}^3}}}} \right]\\
= \dfrac{{72}}{{{{\left( {2x + 1} \right)}^4}}}.\dfrac{{{{\left( {2x + 1} \right)}^3}}}{{ - 12}} = - \dfrac{6}{{2x + 1}}
\end{array}\)
