Đáp án:
$\begin{array}{l}
1)Dkxd:2y - 1 > 0 \Leftrightarrow y > \dfrac{1}{2}\\
\Leftrightarrow \left\{ \begin{array}{l}
3x + \dfrac{1}{{\sqrt {2y - 1} }} = \dfrac{{19}}{3}\\
2x + \dfrac{3}{{\sqrt {2y - 1} }} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
9x + \dfrac{3}{{\sqrt {2y - 1} }} = 19\\
2x + \dfrac{3}{{\sqrt {2y - 1} }} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
7x = 16\\
\dfrac{1}{{\sqrt {2y - 1} }} = \dfrac{{19}}{3} - 3.x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{16}}{7}\\
\dfrac{1}{{\sqrt {2y - 1} }} = \dfrac{{19}}{3} - 3.\dfrac{{16}}{7} = \dfrac{{ - 11}}{{21}}\left( {ktm} \right)
\end{array} \right.
\end{array}$
Vậy hệ pt vô nghiệm
$\begin{array}{l}
2)a){x^2} + \left( {m + 2} \right).x - m - 4 = 0\\
\Delta = {\left( {m + 2} \right)^2} - 4.\left( { - m - 4} \right)\\
= {m^2} + 4m + 4 + 4m + 16\\
= {m^2} + 8m + 16 + 4\\
= {\left( {m + 4} \right)^2} + 4 > 0
\end{array}$
Vậy pt luôn có 2 nghiệm phân biệt
$\begin{array}{l}
b)Khi:{x_1} < 0 \le {x_2}\\
\Leftrightarrow {x_1}.{x_2} \le 0\\
\Leftrightarrow - m - 4 \le 0\\
\Leftrightarrow m \ge - 4\\
Vậy\,m \ge - 4
\end{array}$
