Đáp án:
$\begin{array}{l}
3)\\
a)m = 0\\
\Rightarrow {x^2} - 3x = 0\\
\Rightarrow x\left( {x - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\\
b)\Delta = {\left( {2m + 3} \right)^2} - 4.\left( {{m^2} + 3m} \right)\\
= 4{m^2} + 12m + 9 - 4{m^2} - 12m\\
= 9 > 0\forall m
\end{array}$
=> pt luôn có 2 nghiệm phân biệt
c) Để pt có nghiệm âm thì:
$\begin{array}{l}
\left[ \begin{array}{l}
{x_1}{x_2} < 0\\
\left\{ \begin{array}{l}
{x_1}{x_2} > 0\\
{x_1} + {x_2} < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{m^2} + 3m < 0\\
\left\{ \begin{array}{l}
{m^2} + 3m > 0\\
2m + 3 < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
- 3 < m < 0\\
\left\{ \begin{array}{l}
m < - \dfrac{3}{2}\\
\left[ \begin{array}{l}
m > 0\\
m < - 3
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
- 3 < m < 0\\
m < - 3
\end{array} \right.\\
\Rightarrow m < 0;m \ne - 3\\
4)\\
\left\{ \begin{array}{l}
x + y - 1 = \dfrac{1}{3}\left( {x + 1} \right)\left( {y + 1} \right)\\
x + y + xy = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3\left( {x + y} \right) - 3 = xy + \left( {x + y} \right) + 1\\
x + y + xy = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2\left( {x + y} \right) - xy = 4\\
x + y + xy = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + y = 4\\
xy = 4
\end{array} \right.\\
\Rightarrow x = y = 2
\end{array}$
