Đáp án:
$\begin{array}{l}
a)f\left( x \right) = 3x\left( {{x^2} + 2x} \right) + x - 2{x^2} - 7\\
= 3{x^3} + 6{x^2} + x - 2{x^2} - 7\\
= 3{x^3} + 4{x^2} + x - 7\\
\Rightarrow bac:3\\
g\left( x \right) = 3{x^3} - \left( {2{x^2} - 5x} \right) + 7{x^2} + 3\\
= 3{x^3} - 2{x^2} + 5x + 7{x^2} + 3\\
= 3{x^3} + 5{x^2} + 5x + 3\\
\Rightarrow bac:3\\
b)M\left( x \right) = 2f\left( x \right) + g\left( x \right)\\
= 2.\left( {3{x^3} + 4{x^2} + x - 7} \right) + 3{x^3} + 5{x^2} + 5x + 3\\
= 9{x^3} + 13{x^2} + 7x - 11\\
N\left( x \right) = g\left( x \right) - f\left( x \right)\\
= {x^2} + 4x + 10\\
c){x^2} - 3x = 0\\
\Rightarrow x\left( {x - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0 \Rightarrow M\left( x \right) = - 11\\
x = 3 \Rightarrow M\left( x \right) = 370
\end{array} \right.\\
d)N\left( x \right) = {x^2} + 4x + 10\\
= {x^2} + 4x + 4 + 6\\
= {\left( {x + 2} \right)^2} + 6 \ge 6\forall x\\
\Rightarrow GTNN:N\left( x \right) = 6\\
13)x = 1\\
a)1 - 8.1 + a = 0\\
\Rightarrow a = 7\\
b)1 + a - 5 = 0\\
\Rightarrow a = 4\\
c)a - 1 - 1 = 0\\
\Rightarrow a = 2\\
d)7 + a - 1 = 0\\
\Rightarrow a = - 6
\end{array}$
