Giải hệ phương trình mx+2y=1 { (với m là tham số) 3x+(m+1)y=-1 01/08/2021 Bởi Daisy Giải hệ phương trình mx+2y=1 { (với m là tham số) 3x+(m+1)y=-1
Giải thích các bước giải: Xét m=0 -> \(\left\{ \begin{array}{l}0.x + 2y = 1\\3x + (0 + 1)y = – 1\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}y = \frac{1}{2}\\3x = – 1 – \frac{1}{2} = \frac{{ – 3}}{2}\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}y = \frac{1}{2}\\x = \frac{{ – 1}}{2}\end{array} \right.\) Xét m=-1 \(\left\{ \begin{array}{l} – 1.x + 2y = 1\\3x + ( – 1 + 1)y = – 1\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = \frac{{ – 1}}{3}\\2y = 1 – \frac{1}{3} = \frac{2}{3}\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = \frac{{ – 1}}{3}\\y = \frac{1}{3}\end{array} \right.\) Xét m$\neq$ 0 và -1 \(\begin{array}{l}\left\{ \begin{array}{l}m.x + 2y = 1\\3x + (m + 1)y = – 1\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = \frac{{1 – 2y}}{m}\\3.\frac{{1 – 2y}}{m} + (m + 1)y = – 1\end{array} \right.\\ \leftrightarrow \left\{ \begin{array}{l}x = \frac{{1 – 2y}}{m}\\\frac{3}{m} – \frac{6}{m}.y + (m + 1)y = – 1\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = \frac{{1 – 2y}}{m}\\(m + 1 – \frac{6}{m})y = – 1 – \frac{3}{m}\end{array} \right.\\ \leftrightarrow \left\{ \begin{array}{l}x = \frac{{1 – 2y}}{m}\\\frac{{{m^2} + m – 6}}{m}y = \frac{{ – 3 – m}}{m}\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = \frac{{1 – 2y}}{m}\\\frac{{(m – 2)(m + 3)}}{m}y = \frac{{ – 3 – m}}{m}\end{array} \right.\\ \leftrightarrow \left\{ \begin{array}{l}x = \frac{{1 – 2y}}{m}\\y = \frac{1}{{2 – m}}\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = \frac{{1 – 2.\frac{1}{{2 – m}}}}{m} = \frac{{\frac{{2 – m – 2}}{{2 – m}}}}{m}\\y = \frac{1}{{2 – m}}\end{array} \right. = \frac{1}{{m – 2}}\end{array}\) Bình luận
Đáp án:
Giải thích các bước giải:
Giải thích các bước giải:
Xét m=0
-> \(\left\{ \begin{array}{l}
0.x + 2y = 1\\
3x + (0 + 1)y = – 1
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
y = \frac{1}{2}\\
3x = – 1 – \frac{1}{2} = \frac{{ – 3}}{2}
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
y = \frac{1}{2}\\
x = \frac{{ – 1}}{2}
\end{array} \right.\)
Xét m=-1
\(\left\{ \begin{array}{l}
– 1.x + 2y = 1\\
3x + ( – 1 + 1)y = – 1
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = \frac{{ – 1}}{3}\\
2y = 1 – \frac{1}{3} = \frac{2}{3}
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = \frac{{ – 1}}{3}\\
y = \frac{1}{3}
\end{array} \right.\)
Xét m$\neq$ 0 và -1
\(\begin{array}{l}
\left\{ \begin{array}{l}
m.x + 2y = 1\\
3x + (m + 1)y = – 1
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = \frac{{1 – 2y}}{m}\\
3.\frac{{1 – 2y}}{m} + (m + 1)y = – 1
\end{array} \right.\\
\leftrightarrow \left\{ \begin{array}{l}
x = \frac{{1 – 2y}}{m}\\
\frac{3}{m} – \frac{6}{m}.y + (m + 1)y = – 1
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = \frac{{1 – 2y}}{m}\\
(m + 1 – \frac{6}{m})y = – 1 – \frac{3}{m}
\end{array} \right.\\
\leftrightarrow \left\{ \begin{array}{l}
x = \frac{{1 – 2y}}{m}\\
\frac{{{m^2} + m – 6}}{m}y = \frac{{ – 3 – m}}{m}
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = \frac{{1 – 2y}}{m}\\
\frac{{(m – 2)(m + 3)}}{m}y = \frac{{ – 3 – m}}{m}
\end{array} \right.\\
\leftrightarrow \left\{ \begin{array}{l}
x = \frac{{1 – 2y}}{m}\\
y = \frac{1}{{2 – m}}
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = \frac{{1 – 2.\frac{1}{{2 – m}}}}{m} = \frac{{\frac{{2 – m – 2}}{{2 – m}}}}{m}\\
y = \frac{1}{{2 – m}}
\end{array} \right. = \frac{1}{{m – 2}}
\end{array}\)