Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}y' = \dfrac{{ - \left( {{x^2} - 1} \right)'}}{{{{\left( {{x^2} - 1} \right)}^2}}} = \dfrac{{ - 2x}}{{{{\left( {{x^2} - 1} \right)}^2}}}\\y'' = \dfrac{{ - 2{{\left( {{x^2} - 1} \right)}^2} + 2x.2\left( {{x^2} - 1} \right).2x}}{{{{\left( {{x^2} - 1} \right)}^4}}}\\\,\,\,\,\,\, = \dfrac{{ - 2\left( {{x^2} - 1} \right)\left[ {\left( {{x^2} - 1} \right) - 4{x^2}} \right]}}{{{{\left( {{x^2} - 1} \right)}^4}}}\\\,\,\,\,\,\, = \dfrac{{2\left( {3{x^2} + 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^3}}}\\{y^{\left( 3 \right)}} = 2\dfrac{{6x{{\left( {{x^2} - 1} \right)}^3} - \left( {3{x^2} + 1} \right)3{{\left( {{x^2} - 1} \right)}^2}.2x}}{{{{\left( {{x^2} - 1} \right)}^6}}}\\\,\,\,\,\,\,\, = 2\dfrac{{6x{{\left( {{x^2} - 1} \right)}^2}\left[ {{x^2} - 1 - 3{x^2} - 1} \right]}}{{{{\left( {{x^2} - 1} \right)}^6}}}\\\,\,\,\,\,\, = \dfrac{{12x\left( { - 2{x^2} - 2} \right)}}{{{{\left( {{x^2} - 1} \right)}^4}}}\\ \Rightarrow {y^{\left( 3 \right)}}\left( 2 \right) = \dfrac{{12.2\left( { - {{2.2}^2} - 2} \right)}}{{{{\left( {{2^2} - 1} \right)}^4}}} = \dfrac{{ - 240}}{{81}} = \dfrac{{ - 80}}{{27}}\end{array}\)
Chọn D.
