cho hệ pt x - (m+1)y = 1 4x - y = -2 tìm m nguyên để pt có No duy nhất (x;y) sao cho x y nguyên
1 câu trả lời
Đáp án:
\(m = - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x - \left( {m + 1} \right)y = - 1\\
4x - y = - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4x - 4\left( {m + 1} \right)y = - 4\\
4x - y = - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 4\left( {m + 1} \right)y + y = - 2\\
x = \dfrac{{y - 2}}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( { - 4m - 4 + 1} \right)y = - 2\\
x = \dfrac{{y - 2}}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{2}{{ - 4m - 3}} = \dfrac{2}{{4m + 3}}\\
x = \dfrac{{\dfrac{2}{{4m + 3}} - 2}}{4} = \dfrac{{2 - 8m - 6}}{{4\left( {4m + 3} \right)}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{2}{{4m + 3}}\\
x = \dfrac{{ - 4 - 8m}}{{4\left( {4m + 3} \right)}} = \dfrac{{ - 1 - 2m}}{{4m + 3}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{2}{{4m + 3}}\\
2x = - \dfrac{{4m + 2}}{{4m + 3}} = - \dfrac{{4m + 3 - 1}}{{4m + 3}} = - 1 + \dfrac{1}{{4m + 3}}
\end{array} \right.\\
DK:m \ne - \dfrac{3}{4}\\
Do:x \in Z;y \in Z\\
\to \left\{ \begin{array}{l}
\dfrac{2}{{4m + 3}} \in Z\\
\dfrac{1}{{4m + 3}} \in Z
\end{array} \right.\\
\to \dfrac{1}{{4m + 3}} \in Z\\
\to 4m + 3 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
4m + 3 = 1\\
4m + 3 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = - \dfrac{1}{2}\left( {KTM} \right)\\
m = - 1
\end{array} \right.\left( {do:m \in Z} \right)\\
\to m = - 1
\end{array}\)