Đáp án:
$\begin{array}{l}
a){a^2} + 9{b^2} = 8ab\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} + 6ab + 9{b^2} = 8ab + 6ab\\
{a^2} - 6ab + 9{b^2} = 8ab - 6ab
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {a + 3b} \right)^2} = 14ab\\
{\left( {a - 3b} \right)^2} = 2ab
\end{array} \right.\\
\Rightarrow P = \frac{{{{\left( {a + 3b} \right)}^2}}}{{{{\left( {a - 3b} \right)}^2}}} = \frac{{14ab}}{{2ab}} = 7\\
b){a^2} + {b^2} + ab = 7\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} + {b^2} = 7 - ab\\
{a^2} + {b^2} + 2ab = 7 + ab
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} + {b^2} = 7 - ab\\
{\left( {a + b} \right)^2} = 7 + ab
\end{array} \right.\left( 1 \right)\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {{a^2} + {b^2}} \right)^2} = {\left( {7 - ab} \right)^2} \Rightarrow {a^4} + {b^4} + 2{a^2}{b^2} = 49 - 14ab + {a^2}{b^2}\\
{\left( {a + b} \right)^4} = {\left( {7 + ab} \right)^2} = 49 + {a^2}{b^2} + 14ab
\end{array} \right.\\
\Rightarrow {a^4} + {b^4} = 49 - {a^2}{b^2} - 14ab\\
\Rightarrow A = \frac{{{a^2} + {b^2} + {{\left( {a + b} \right)}^2}}}{{{a^4} + {b^4} + {{\left( {a + b} \right)}^4}}}\\
= \frac{{7 - ab + 7 + ab}}{{49 - {a^2}{b^2} - 14ab + 49 + {a^2}{b^2} + 14ab}}\\
= \frac{{14}}{{98}} = \frac{1}{7}
\end{array}$
