cho hệ pt :x+2y=5,mx+y=4 a)Tìm m để hệ pt có nghiệm duy nhất mà x vào y trái dấu b)Tìm m để hệ pt có nghiệm duy nhất mà x=IyI

Đáp án:

$\begin{array}{l}
a)\left\{ \begin{array}{l}
x + 2y = 5\\
mx + y = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 2y = 5\\
2mx + 2y = 8
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2mx - x = 3\\
y = 4 - mx
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {2m - 1} \right).x = 3\\
y = 4 - mx
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{3}{{2m - 1}}\left( {m \ne \dfrac{1}{2}} \right)\\
y = 4 - m.\dfrac{3}{{2m - 1}} = \dfrac{{5m - 4}}{{2m - 1}}
\end{array} \right.\\
 + khi:x.y < 0\\
 \Leftrightarrow \dfrac{3}{{2m - 1}}.\dfrac{{5m - 4}}{{2m - 1}} < 0\\
 \Leftrightarrow \dfrac{{3.\left( {5m - 4} \right)}}{{{{\left( {2m - 1} \right)}^2}}} < 0\\
 \Leftrightarrow 5m - 4 < 0\\
 \Leftrightarrow m < \dfrac{4}{5}\\
Vậy\,m < \dfrac{4}{5};m \ne \dfrac{1}{2}\\
b)m \ne \dfrac{1}{2}\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{3}{{2m - 1}}\\
y = \dfrac{{5m - 4}}{{2m - 1}}
\end{array} \right.\\
x = \left| y \right|\\
 \Leftrightarrow \dfrac{3}{{2m - 1}} = \left| {\dfrac{{5m - 4}}{{2m - 1}}} \right|\\
 \Leftrightarrow \left\{ \begin{array}{l}
2m - 1 > 0\\
\left[ \begin{array}{l}
\dfrac{3}{{2m - 1}} = \dfrac{{5m - 4}}{{2m - 1}}\\
\dfrac{3}{{2m - 1}} = \dfrac{{4 - 5m}}{{2m - 1}}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m > \dfrac{1}{2}\\
\left[ \begin{array}{l}
5m - 4 = 3\\
4 - 5m = 3
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m > \dfrac{1}{2}\\
\left[ \begin{array}{l}
m = \dfrac{7}{5}\left( {tm} \right)\\
m = \dfrac{1}{5}\left( {ktm} \right)
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow m = \dfrac{7}{5}\\
Vậy\,m = \dfrac{7}{5}
\end{array}$