Đáp án:
$\begin{array}{l}
Do:{\left( {4a + 5b} \right)^6} \ge 0\forall a,b\\
\left| {7{a^2} + 8{b^2} - 303} \right| \ge 0\forall a,b\\
\Rightarrow {\left| {7{a^2} + 8{b^2} - 303} \right|^9} \ge 0\\
\Rightarrow {\left( {4a + 5b} \right)^6} + {\left| {7{a^2} + 8{b^2} - 303} \right|^9} \ge 0\\
Do:{\left( {4a + 5b} \right)^6} + {\left| {7{a^2} + 8{b^2} - 303} \right|^9} \le 0
\end{array}$
=> Dấu = sẽ xảy ra
$\begin{array}{l}
\Rightarrow \left\{ \begin{array}{l}
4a + 5b = 0\\
7{a^2} + 8{b^2} - 303 = 0
\end{array} \right.\\
\Rightarrow \left\{ {a = - \dfrac{5}{4}b} \right.\\
\Rightarrow 7.{\left( { - \dfrac{5}{4}b} \right)^2} + 8{b^2} = 303\\
\Rightarrow \dfrac{{303{b^2}}}{{16}} = 303\\
\Rightarrow {b^2} = 16\\
\Rightarrow \left[ \begin{array}{l}
b = 4\\
b = - 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
a = - \dfrac{5}{4}b = - 5\\
a = - \dfrac{5}{4}b = 5
\end{array} \right.\\
Vậy\,\left( {a;b} \right) = \left( { - 5;4} \right)/\left( {5; - 4} \right)
\end{array}$
