1) TH1: m = 0, bpt trở thành \(- 6x + 14 \ge 0 \Leftrightarrow x \le \frac{7}{3}\) (không thỏa ycbt).

TH2: \(m \ne 0\),  \(m{x^2} - 2\left( {m + 3} \right)x + 2m + 14 \ge 0\) vô nghiệm \( \Leftrightarrow m{x^2} - 2\left( {m + 3} \right)x + 2m + 14 < 0\) có nghiệm \(\forall x \in R\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{m < 0}\\
{\Delta ' < 0}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{m < 0}\\
{ - {m^2} - 8m + 9 < 0}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{m < 0}\\
{m <  - 9\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{ \vee m > 1}
\end{array}}
\end{array}}
\end{array}} \right. \Leftrightarrow m <  - 9.\)

Vậy m < - 9.

2) TH1: \({x^2} - 5x + 6 = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\)

TH2:  \({x^2} - 5x + 6 > 0 \Leftrightarrow \left[ \begin{array}{l}
x < 2\\
x > 3
\end{array} \right.\) Khi đó, bpt \( \Leftrightarrow \sqrt {2{x^2} + 4}  \ge x + 2 \Leftrightarrow \left[ \begin{array}{l}
x + 2 < 0\\
\left\{ \begin{array}{l}
x + 2 \ge 0\\
2{x^2} + 4 \ge {\left( {x + 2} \right)^2}
\end{array} \right.
\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}
x <  - 2\\
\left\{ \begin{array}{l}
x \ge  - 2\\
{x^2} - 4x \ge 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x <  - 2\\
\left\{ \begin{array}{l}
x \ge  - 2\\
x \ge 4 \vee x \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \le 0\\
x \ge 4
\end{array} \right.\)

Vậy tập nghiệm bất phương trình \(S = \left( { - \infty ;0} \right] \cup \left\{ {2,3} \right\} \cup \left[ {4; + \infty } \right)\)

3) Hpt : \(\left\{ \begin{array}{l}
{x^2} + {x^3}y - x{y^2} + xy - y = 1\\
{x^4} + {y^2} - xy\left( {2x - 1} \right) = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - y + xy\left( {{x^2} - y} \right) + xy = 1\\
{\left( {{x^2} - y} \right)^2} + xy = 1
\end{array} \right.\)

Đặt \(a = {x^2} - y,b = xy\) hệ thành :

\(\left\{ \begin{array}{l}
a + ab + b = 1\\
{a^2} + b = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{a^3} + {a^2} - 2a = 0\\
b = 1 - {a^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 0\\
b = 1
\end{array} \right. \vee \left\{ \begin{array}{l}
a = 1\\
b = 0
\end{array} \right. \vee \left\{ \begin{array}{l}
a =  - 2\\
b =  - 3
\end{array} \right..\)

+ Với \(\left\{ \begin{array}{l}
a = 0\\
b = 1
\end{array} \right.\) ta có \(\left\{ \begin{array}{l}
{x^2} - y = 0\\
xy = 1
\end{array} \right. \Leftrightarrow x = y = 1.\)

+ Với \(\left\{ \begin{array}{l}
a = 1\\
b = 0
\end{array} \right.\) ta có \(\left\{ \begin{array}{l}
{x^2} - y = 1\\
xy = 0
\end{array} \right. \Leftrightarrow \left( {x;y} \right) \in \left\{ {\left( {0; - 1} \right),\left( {1;0} \right),\left( { - 1;0} \right)} \right\}.\)

+ Với \(\left\{ \begin{array}{l}
a =  - 2\\
b =  - 3
\end{array} \right.\) ta có \(\left\{ \begin{array}{l}
{x^2} - y =  - 2\\
xy =  - 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y =  - \frac{3}{x}\\
\left( {x + 1} \right)\left( {{x^2} - x + 3} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
y = 3
\end{array} \right..\)

Vậy hệ có 5 nghiệm \(\left( {x;y} \right) \in \left\{ {\left( {1;1} \right),\left( {0; - 1} \right),\left( {1;0} \right),\left( { - 1;0} \right),\left( { - 1;3} \right)} \right\}.\)