Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{{x^4} + 7{x^2} + 4}}{{{x^2} + 2}} = 4x\\
\to \frac{{{x^4} + 7{x^2} + 4 - 4{x^3} - 8x}}{{{x^2} + 2}} = 0\\
\to {x^4} + 7{x^2} + 4 - 4{x^3} - 8x = 0\\
\to {x^4} - {x^3} - 3{x^3} + 3{x^2} + 4{x^2} - 4x - 4x + 4 = 0\\
\to {x^3}\left( {x - 1} \right) - 3{x^2}\left( {x - 1} \right) + 4x\left( {x - 1} \right) - 4\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^3} - 3{x^2} + 4x - 4} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^3} - 2{x^2} - {x^2} + 2x + 2x - 4} \right) = 0\\
\to \left( {x - 1} \right)\left( {x - 2} \right)\left( {{x^2} - x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
x - 2 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)
