b) bỏ x = | y| Tìm nghiệm pb thôi

Đáp án:

$\begin{array}{l}
a)m = 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + y = 1 + 1\\
x + y = 3.1 - 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + y = 2\\
x + y = 2
\end{array} \right.\left( {tm} \right)\\
Vay\,x + y = 2\\
b)\left\{ \begin{array}{l}
x + my = m + 1\\
mx + y = 3m - 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
mx + {m^2}y = {m^2} + m\\
mx + y = 3m - 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{m^2}y - y = {m^2} + m - 3m + 1\\
x = m + 1 - my
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {{m^2} - 1} \right).y = {m^2} - 2m + 1\\
x = m + 1 - my
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {m - 1} \right)\left( {m + 1} \right).y = {\left( {m - 1} \right)^2}\\
x = m + 1 - my
\end{array} \right.
\end{array}$

Hệ có nghiệm duy nhất thì:

$\begin{array}{l}
\left( {m - 1} \right)\left( {m + 1} \right) \ne 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
m \ne  - 1
\end{array} \right.\\
 \Leftrightarrow y = \dfrac{{{{\left( {m - 1} \right)}^2}}}{{\left( {m - 1} \right)\left( {m + 1} \right)}} = \dfrac{{m - 1}}{{m + 1}}\\
 \Leftrightarrow x = m + 1 - my = m + 1 - m.\dfrac{{m - 1}}{{m + 1}} = \dfrac{{3m + 1}}{{m + 1}}\\
Khi:x = \left| y \right|\\
 \Leftrightarrow \dfrac{{3m + 1}}{{m + 1}} = \left| {\dfrac{{m - 1}}{{m + 1}}} \right|\left( {dk:m \ge  - \dfrac{1}{3}/m <  - 1} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3m + 1 = m - 1\\
3m + 1 = 1 - m
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2m =  - 2\\
4m = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
m =  - 1\left( {ktm} \right)\\
m = 0\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = 0
\end{array}$