Đáp án:
$\begin{array}{l}
3)M = \sqrt {3x - 5} - \frac{1}{{\sqrt[3]{{{x^2} - 4}}}}\\
\Rightarrow Dkxd:\left\{ \begin{array}{l}
3x - 5 \ge 0\\
{x^2} - 4 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \ge \frac{5}{3}\\
x \ne 2;x \ne - 2
\end{array} \right.\\
\Rightarrow x \ge \frac{5}{3};x \ne 2\\
1)Theo\,Pytago:\\
M{P^2} = M{N^2} + N{P^2} = 25{a^2}\\
\Rightarrow MP = 5a
\end{array}$
Khi quay tam giác MNP quanh MN thì được hình nón có bán kính đáy NP, đường sinh MN, chiều cao MN
$\begin{array}{l}
\Rightarrow {S_{xq}} = \pi .NP.MP = \pi .3a.5a = 15\pi .{a^2}\\
2)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
{x_1}{x_2} = 1
\end{array} \right.\\
{t_1} = 2{x_1} - x_2^2;{t_2} = 2{x_2} - x_1^2\\
\Rightarrow \left\{ \begin{array}{l}
{t_1} + {t_2} = 2\left( {{x_1} + {x_2}} \right) - \left( {x_1^2 + x_2^2} \right)\\
{t_1}{t_2} = \left( {2{x_1} - x_2^2} \right)\left( {2{x_2} - x_1^2} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{t_1} + {t_2} = 2.3 - {\left( {{x_1} + {x_2}} \right)^2} + 2{x_1}{x_2}\\
{t_1}{t_2} = 4{x_1}{x_2} - 2\left( {x_1^3 + x_2^3} \right) + x_1^2x_2^2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{t_1} + {t_2} = 2.3 - {3^2} + 2.1\\
{t_1}{t_2} = 4.1 - 2\left[ {{3^3} - 3.1.3} \right] + {1^1}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{t_1} + {t_2} = - 1\\
{t_1}{t_2} = - 31
\end{array} \right.\\
\Rightarrow Pt:{t^2} + t - 31 = 0
\end{array}$
