Đáp án đúng: A
Giải chi tiết:Đặt \(x = {a^2},y = {b^2}\) thì \(M = \frac{{x + y}}{{x - y}} + \frac{{x - y}}{{x + y}} = \frac{{{{\left( {x + y} \right)}^2} + {{\left( {x - y} \right)}^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \frac{{2\left( {{x^2} + {y^2}} \right)}}{{{x^2} - {y^2}}}\)
\(N = \frac{{{x^4} + {y^4}}}{{{x^4} - {y^4}}} + \frac{{{x^4} - {y^4}}}{{{x^4} + {y^4}}}\)
Ta có:
\(\begin{array}{l}\frac{{{x^4} + {y^4}}}{{{x^4} - {y^4}}} = \frac{1}{2}.\frac{{{{\left( {{x^2} + {y^2}} \right)}^2} + {{\left( {{x^2} - {y^2}} \right)}^2}}}{{\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {\frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}} + \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {\frac{M}{2} + \frac{2}{M}} \right) = \frac{{{M^2} + 4}}{{4M}}\end{array}\)
\( \Rightarrow \frac{{{x^4} - {y^4}}}{{{x^4} + {y^4}}} = \frac{{4M}}{{{M^2} + 4}}\)
\( \Rightarrow N = \frac{{{M^2} + 4}}{{4M}} + \frac{{4M}}{{{M^2} + 4}} = \frac{{{{\left( {{M^2} + 4} \right)}^2} + 16{M^2}}}{{4M\left( {{M^2} + 4} \right)}} = \frac{{{M^4} + 24{M^2} + 16}}{{4M\left( {{M^2} + 4} \right)}}\).
Vậy \(N = \frac{{{M^4} + 24{M^2} + 16}}{{4M\left( {{M^2} + 4} \right)}}.\)
Chọn A.