Đáp án:
b) \(x = \dfrac{{a - 3}}{a}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){a^2}x + x = 2{a^4} - 2\\
\to \left( {{a^2} + 1} \right)x = 2\left( {{a^2} - 1} \right)\left( {{a^2} + 1} \right)\\
\to x = \dfrac{{2\left( {{a^2} - 1} \right)\left( {{a^2} + 1} \right)}}{{{a^2} + 1}}\\
\to x = 2\left( {{a^2} - 1} \right)\\
b){a^2}x + 3ax + 9 = {a^2}\\
\to \left( {{a^2} + 3a} \right)x = {a^2} - 9\\
\to a\left( {a + 3} \right)x = \left( {a - 3} \right)\left( {a + 3} \right)\\
\to x = \dfrac{{\left( {a - 3} \right)\left( {a + 3} \right)}}{{a\left( {a + 3} \right)}}\\
\to x = \dfrac{{a - 3}}{a}
\end{array}\)
