Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\,\,\,\,\,{x^2} - 2 \ne 0 \Leftrightarrow x \ne \pm \sqrt 2 \\
\frac{{{x^4} + 4}}{{{x^2} - 2}} = 5x\\
\Leftrightarrow {x^4} + 4 = 5x.\left( {{x^2} - 2} \right)\\
\Leftrightarrow {x^4} + 4 = 5{x^3} - 10x\\
\Leftrightarrow {x^4} - 5{x^3} + 10x + 4 = 0\\
\Leftrightarrow \left( {{x^4} + {x^3}} \right) - \left( {6{x^3} + 6{x^2}} \right) + \left( {6{x^2} + 6x} \right) + \left( {4x + 4} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^3} - 6{x^2} + 6x + 4} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left[ {\left( {{x^3} - 2{x^2}} \right) - \left( {4{x^2} - 8x} \right) - \left( {2x - 4} \right)} \right] = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x - 2} \right)\left( {{x^2} - 4x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x - 2 = 0\\
{x^2} - 4x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 2\\
x = 2 \pm \sqrt 6
\end{array} \right.\\
b,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne - 9\\
{x^2} + \frac{{81{x^2}}}{{{{\left( {x + 9} \right)}^2}}} = 40\\
\Leftrightarrow {x^2}{\left( {x + 9} \right)^2} + 81{x^2} = 40{\left( {x + 9} \right)^2}\\
\Leftrightarrow {x^2}\left( {{x^2} + 18x + 81} \right) + 81{x^2} = 40\left( {{x^2} + 18x + 81} \right)\\
\Leftrightarrow {x^4} + 18{x^3} + 81{x^2} + 81{x^2} = 40{x^2} + 720x + 3240\\
\Leftrightarrow {x^4} + 18{x^3} + 122{x^2} - 720x - 3240 = 0\\
\Leftrightarrow \left( {{x^4} - 2{x^3} - 18{x^2}} \right) + \left( {20{x^3} - 40{x^2} - 360x} \right) + \left( {180{x^2} - 360x - 3240} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 2x - 18} \right)\left( {{x^2} + 20x + 180} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x - 18 = 0\\
{x^2} + 20x + 180 = 0\,\,\,\,\left( {VN} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1 + \sqrt {19} \\
x = 1 - \sqrt {19}
\end{array} \right.
\end{array}\)
