giải hệ theo m. a/ (m-1)x+y=1 và 2x+(m+1)y=-3 b/ (m+1)x-my=5 và x+my=m^2+4m. Tìm m nguyên để x,y nguyên cảm ơn các bạn 18/07/2021 Bởi Jade giải hệ theo m. a/ (m-1)x+y=1 và 2x+(m+1)y=-3 b/ (m+1)x-my=5 và x+my=m^2+4m. Tìm m nguyên để x,y nguyên cảm ơn các bạn
Giải thích các bước giải: \(\begin{array}{l}a)\left\{ \begin{array}{l}\left( {m – 1} \right)x + y = 1\\2x + \left( {m + 1} \right)y = – 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = 1 – \left( {m – 1} \right)x\\2x + \left( {m – 1} \right)\left( {1 – \left( {m + 1} \right)} \right)x = – 3\end{array} \right.\\ \Rightarrow 2x – \left( {{m^2} – 1} \right)x = – 3 – m + 1\\ \Leftrightarrow \left( {{m^2} – 3} \right)x = m + 2\\TH1:m = \sqrt 3 \Rightarrow 0 = \sqrt 3 + 2\left( {VN} \right)\\TH2:m = – \sqrt 3 \Rightarrow 0 = – \sqrt 3 + 2\left( {VN} \right)\\TH3:m \ne \pm \sqrt 3 \Rightarrow x = \dfrac{{m + 2}}{{{m^2} – 3}}\\ \Rightarrow y = 1 – \left( {m – 1} \right).\dfrac{{m + 2}}{{{m^2} – 3}} = \dfrac{{ – m – 1}}{{{m^2} – 3}}\\b)\left\{ \begin{array}{l}\left( {m + 1} \right)x – my = 5\\x + my = {m^2} + 4m\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = {m^2} + 4m – my\\\left( {m + 1} \right)\left( {{m^2} + 4m – my} \right) – my = 5\end{array} \right.\\ \Rightarrow – m\left( {m + 2} \right)y = 5 – \left( {{m^2} + 4m} \right)\left( {m + 1} \right)\\ \Leftrightarrow m\left( {m + 2} \right)y = {m^3} + 5{m^2} + 4m – 5\\ \Rightarrow y = \dfrac{{{m^3} + 5{m^2} + 4m – 5}}{{{m^2} + 2m}}\left( {m \ne 0;m \ne – 2} \right)\\y = m + 3 – \dfrac{{2m + 5}}{{{m^2} + 2m}}\\y \in Z \Rightarrow \dfrac{{2m + 5}}{{{m^2} + 2m}} \in Z \Rightarrow \dfrac{{2{m^2} + 5m}}{{{m^2} + 2m}} = 2 + \dfrac{1}{{m + 2}} \in Z\\ \Rightarrow m + 2 \in \{ – 1;1\} \\m + 2 = 1 \Rightarrow m = – 1 \Rightarrow y = 5\left( {tm} \right)\\m + 2 = – 1 \Rightarrow m = – 3 \Rightarrow y = \dfrac{1}{3}\left( {ktm} \right)\end{array}\) Bình luận
Giải thích các bước giải:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
\left( {m – 1} \right)x + y = 1\\
2x + \left( {m + 1} \right)y = – 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 1 – \left( {m – 1} \right)x\\
2x + \left( {m – 1} \right)\left( {1 – \left( {m + 1} \right)} \right)x = – 3
\end{array} \right.\\
\Rightarrow 2x – \left( {{m^2} – 1} \right)x = – 3 – m + 1\\
\Leftrightarrow \left( {{m^2} – 3} \right)x = m + 2\\
TH1:m = \sqrt 3 \Rightarrow 0 = \sqrt 3 + 2\left( {VN} \right)\\
TH2:m = – \sqrt 3 \Rightarrow 0 = – \sqrt 3 + 2\left( {VN} \right)\\
TH3:m \ne \pm \sqrt 3 \Rightarrow x = \dfrac{{m + 2}}{{{m^2} – 3}}\\
\Rightarrow y = 1 – \left( {m – 1} \right).\dfrac{{m + 2}}{{{m^2} – 3}} = \dfrac{{ – m – 1}}{{{m^2} – 3}}\\
b)\left\{ \begin{array}{l}
\left( {m + 1} \right)x – my = 5\\
x + my = {m^2} + 4m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = {m^2} + 4m – my\\
\left( {m + 1} \right)\left( {{m^2} + 4m – my} \right) – my = 5
\end{array} \right.\\
\Rightarrow – m\left( {m + 2} \right)y = 5 – \left( {{m^2} + 4m} \right)\left( {m + 1} \right)\\
\Leftrightarrow m\left( {m + 2} \right)y = {m^3} + 5{m^2} + 4m – 5\\
\Rightarrow y = \dfrac{{{m^3} + 5{m^2} + 4m – 5}}{{{m^2} + 2m}}\left( {m \ne 0;m \ne – 2} \right)\\
y = m + 3 – \dfrac{{2m + 5}}{{{m^2} + 2m}}\\
y \in Z \Rightarrow \dfrac{{2m + 5}}{{{m^2} + 2m}} \in Z \Rightarrow \dfrac{{2{m^2} + 5m}}{{{m^2} + 2m}} = 2 + \dfrac{1}{{m + 2}} \in Z\\
\Rightarrow m + 2 \in \{ – 1;1\} \\
m + 2 = 1 \Rightarrow m = – 1 \Rightarrow y = 5\left( {tm} \right)\\
m + 2 = – 1 \Rightarrow m = – 3 \Rightarrow y = \dfrac{1}{3}\left( {ktm} \right)
\end{array}\)