Đáp án:
2) m>2
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
2x - 3y = 5m - 11\\
5x + 4y = m + 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
8x - 12y = 20m - 44\\
15x + 12y = 3m + 21
\end{array} \right.\\
\to \left\{ \begin{array}{l}
23x = 23m - 23\\
y = \dfrac{{2x - 5m + 11}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m - 1\\
y = \dfrac{{2m - 2 - 5m + 11}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m - 1\\
y = - m + 3
\end{array} \right.\\
1){x^2} + {y^2} = 10\\
\to {\left( {m - 1} \right)^2} + {\left( {3 - m} \right)^2} = 10\\
\to {m^2} - 2m + 1 + 9 - 6m + {m^2} = 10\\
\to 2{m^2} - 8m = 0\\
\to \left[ \begin{array}{l}
m = 0\\
m = 4
\end{array} \right.\\
2)2x - 5y > - 3\\
\to 2\left( {m - 1} \right) - 5\left( {3 - m} \right) > - 3\\
\to 2m - 2 - 15 + 5m > - 3\\
\to 7m > 14\\
\to m > 2\\
3)\left\{ \begin{array}{l}
x > 0\\
y < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m - 1 > 0\\
3 - m < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 1\\
m > 3
\end{array} \right.\\
\to m > 3\\
4)\dfrac{2}{x} + \dfrac{3}{y} = - 1\\
\to \dfrac{2}{{m - 1}} + \dfrac{3}{{3 - m}} = - 1\\
\to 6 - 2m + 3m - 3 = \left( {m - 1} \right)\left( {3 - m} \right)\\
\to m + 3 = - {m^2} + 4m - 3\\
\to {m^2} - 3m + 6 = 0\\
Do:{m^2} - 3m + 6 > 0\forall m\\
\to x \in \emptyset \\
5)\left| {2x + 1} \right| = y\\
\to \left[ \begin{array}{l}
2x + 1 = y\\
2x + 1 = - y
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\left( {m - 1} \right) + 1 = 3 - m\\
2\left( {m - 1} \right) + 1 = - 3 + m
\end{array} \right.\\
\to \left[ \begin{array}{l}
3m = 4\\
m = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = \dfrac{4}{3}\\
m = - 2
\end{array} \right.
\end{array}\)
