Trang chủ » Hỏi đáp » Môn Toán cho hệ pt mx +4y=4-m x+my=1 tìm m nguyên để hệ pt có nghiệm duy nhất s cho x<-1,y<-1 02/03/2024 cho hệ pt mx +4y=4-m x+my=1 tìm m nguyên để hệ pt có nghiệm duy nhất s cho x<-1,y<-1
Giải đáp: $ – 4 < m < – 2$ Lời giải và giải thích chi tiết: $\begin{array}{l}\left\{ \begin{array}{l}mx + 4y = 4 – m\\x + my = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}mx + 4y = 4 – m\\mx + {m^2}y = m\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}mx + {m^2}y – mx – 4y = m – \left( {4 – m} \right)\\x + my = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left( {{m^2} – 4} \right)y = m – 4 + m\\x = 1 – my\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left( {m – 2} \right)\left( {m + 2} \right).y = 2m – 4\\x = 1 – my\end{array} \right.\\ \Leftrightarrow m \ne 2;m \ne – 2\\ \Leftrightarrow y = \dfrac{{2m – 4}}{{\left( {m – 2} \right)\left( {m + 2} \right)}} = \dfrac{2}{{m + 2}}\\ \Leftrightarrow x = 1 – my = 1 – m.\dfrac{2}{{m + 2}} = \dfrac{{2 – m}}{{m + 2}}\\ \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{{2 – m}}{{m + 2}}\\y = \dfrac{2}{{m + 2}}\end{array} \right.\\Khi:x < – 1;y < – 1\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{2 – m}}{{m + 2}} < – 1\\\dfrac{2}{{m + 2}} < – 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{2 – m}}{{m + 2}} + 1 < 0\\\dfrac{2}{{m + 2}} + 1 < 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{4}{{m + 2}} < 0\\\dfrac{{m + 4}}{{m + 2}} < 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m + 2 < 0\\m + 4 > 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m < – 2\\m > – 4\end{array} \right.\\ \Leftrightarrow – 4 < m < – 2\end{array}$ Trả lời
\left\{ \begin{array}{l}
mx + 4y = 4 – m\\
x + my = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
mx + 4y = 4 – m\\
mx + {m^2}y = m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
mx + {m^2}y – mx – 4y = m – \left( {4 – m} \right)\\
x + my = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {{m^2} – 4} \right)y = m – 4 + m\\
x = 1 – my
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {m – 2} \right)\left( {m + 2} \right).y = 2m – 4\\
x = 1 – my
\end{array} \right.\\
\Leftrightarrow m \ne 2;m \ne – 2\\
\Leftrightarrow y = \dfrac{{2m – 4}}{{\left( {m – 2} \right)\left( {m + 2} \right)}} = \dfrac{2}{{m + 2}}\\
\Leftrightarrow x = 1 – my = 1 – m.\dfrac{2}{{m + 2}} = \dfrac{{2 – m}}{{m + 2}}\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{2 – m}}{{m + 2}}\\
y = \dfrac{2}{{m + 2}}
\end{array} \right.\\
Khi:x < – 1;y < – 1\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{2 – m}}{{m + 2}} < – 1\\
\dfrac{2}{{m + 2}} < – 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{2 – m}}{{m + 2}} + 1 < 0\\
\dfrac{2}{{m + 2}} + 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{4}{{m + 2}} < 0\\
\dfrac{{m + 4}}{{m + 2}} < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m + 2 < 0\\
m + 4 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < – 2\\
m > – 4
\end{array} \right.\\
\Leftrightarrow – 4 < m < – 2
\end{array}$