cần gấp tìm m để hpt $\left \{ {{x-my=1} \atop {mx+y=-m}} \right.$ có ngh duy nhất (x;y) tm x<1 và y<1
2 câu trả lời
Đáp án:$m \ne 0;m \ne - 1$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
x - my = 1\\
mx + y = - m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
mx - {m^2}y = m\\
mx + y = - m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y + {m^2}y = - m - m\\
x = my + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {{m^2} + 1} \right).y = - 2m\\
x = my + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = \dfrac{{ - 2m}}{{{m^2} + 1}}\\
x = m.\dfrac{{ - 2m}}{{{m^2} + 1}} + 1 = \dfrac{{1 - {m^2}}}{{{m^2} + 1}}
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 1\\
y < 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{1 - {m^2}}}{{{m^2} + 1}} - 1 < 0\\
\dfrac{{ - 2m}}{{{m^2} + 1}} - 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{1 - {m^2} - {m^2} - 1}}{{{m^2} + 1}} < 0\\
\dfrac{{ - 2m - {m^2} - 1}}{{{m^2} + 1}} < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{ - 2{m^2}}}{{{m^2} + 1}} < 0\\
- {m^2} - 2m - 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 2{m^2} < 0\\
{\left( {m + 1} \right)^2} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} > 0\\
{\left( {m + 1} \right)^2} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
m + 1 \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
m \ne - 1
\end{array} \right.\\
Vậy\,m \ne 0;m \ne - 1
\end{array}$