Nếu biết $3{\sin ^4}x + 2{\cos ^4}x = \dfrac{{98}}{{81}}$ thì giá trị biểu thức $A = 2{\sin ^4}x + 3{\cos ^4}x$ bằng

Ta có:

\(\begin{array}{l}3{\sin ^4}x + 2{\cos ^4}x = \dfrac{{98}}{{81}}\\2{\sin ^4}x + 3{\cos ^4}x = A\end{array}\)

Trừ vế cho vế hai đẳng thức trên ta được:

\(\begin{array}{l}{\sin ^4}x - {\cos ^4}x = \dfrac{{98}}{{81}} - A\\ \Leftrightarrow \left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = \dfrac{{98}}{{81}} - A\\ \Leftrightarrow {\sin ^2}x - {\cos ^2}x = \dfrac{{98}}{{81}} - A\\ \Leftrightarrow {\cos ^2}x - {\sin ^2}x = A - \dfrac{{98}}{{81}}\\ \Leftrightarrow \cos 2x = A - \dfrac{{98}}{{81}}\end{array}\)

Cộng vế với vế hai đẳng thức đầu bài ta được:

\(\begin{array}{l}5{\sin ^4}x + 5{\cos ^4}x = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right] = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left( {1 - \dfrac{1}{2}.4{{\sin }^2}x{{\cos }^2}x} \right) = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right) = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 1 - \dfrac{1}{2}{\sin ^2}2x = \dfrac{1}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\\ \Leftrightarrow 1 - \dfrac{1}{2}\left( {1 - {{\cos }^2}2x} \right) = \dfrac{1}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\\ \Leftrightarrow \dfrac{1}{2} + \dfrac{1}{2}{\cos ^2}2x = \dfrac{1}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\\ \Leftrightarrow 1 + {\cos ^2}2x = \dfrac{2}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\end{array}\)

Thay \(\cos 2x = A - \dfrac{{98}}{{81}}\) ta được:

\( 1 + {\left( {A - \dfrac{{98}}{{81}}} \right)^2} = \dfrac{2}{5}\left( {A + \dfrac{{98}}{{81}}} \right) \) \(= \dfrac{2}{5}\left( {A - \dfrac{{98}}{{81}}} \right) + \dfrac{{392}}{{405}}\)

Đặt \(A - \dfrac{{98}}{{81}} = t\)\( \Rightarrow {t^2} - \dfrac{2}{5}t + \dfrac{{13}}{{405}} = 0\)\( \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{{13}}{{45}}\\t = \dfrac{1}{9}\end{array} \right.\)

+) \(t = \dfrac{{13}}{{45}} \Rightarrow A = \dfrac{{607}}{{405}}\)

+) \(t = \dfrac{1}{9} \Rightarrow A = \dfrac{{107}}{{81}}.\)