$\left \{ {{mx + y =2m} \atop {x+my=m+1}} \right.$ Tìm các giá trị nguyên của m để hpt có nghiệm x, y là các số nguyên

$\begin{cases} mx+y=2m\\ x+my=m+1 \end{cases}\Leftrightarrow \begin{cases} x=m+1-my\\ m(m+1-my)+y=2m \end{cases}\\Leftrightarrow \begin{cases} x=m+1-my\\ y(1-m^2)=m-m^2 \end{cases}\Leftrightarrow\begin{cases} x=m+1-\dfrac{m^2}{m+1}\\ y=\dfrac{m}{m+1} \end{cases}\Leftrightarrow\begin{cases} x=\dfrac{m^2+2m+1-m^2}{m+1}=\dfrac{2m+1}{m+1}\\ y=\dfrac{m}{m+1} \end{cases}$

Ta có: $m\neq -1$

Để : $x,y\in \mathbb{Z}\Leftrightarrow \dfrac{2m+1}{m+1}$ và $\dfrac{m}{m+1}\in \mathbb{Z}$

Xét $\dfrac{2m+1}{m+1}=2-\dfrac{1}{m+1}\in \mathbb{Z}$

$\Leftrightarrow m+1\in Ư(1)=\left\{-1;1\right\}$

$\Leftrightarrow m\in \left\{-2;0\right\}(a)$

Xét $\dfrac{m}{m+1}=1-\dfrac{1}{m+1}\in \mathbb{Z}$

$\Leftrightarrow m+1\in Ư(1)=\left\{-1;1\right\}$

$\Leftrightarrow m\in \left\{-2;0\right\}(b)$

Từ $(a)(b)\Rightarrow m\in \left\{-2;0\right\}$

Vậy $m\in \left\{-2;0\right\}$.

$\begin{cases} mx+y=2m\\ x+my=m+1 \end{cases}\Leftrightarrow \begin{cases} x=m+1-my\\ m(m+1-my)+y=2m \end{cases}\Leftrightarrow \begin{cases} x=m+1-my\\ m^2+m-m^2y+y=2m \end{cases}\Leftrightarrow \begin{cases} x=m+1-my\\ y(1-m^2)=m-m^2 \end{cases}\Leftrightarrow \begin{cases} x=m+1-my\\ y=\dfrac{m(1-m)}{(1-m)(1+m)}=\dfrac{m}{m+1} \end{cases}\Leftrightarrow \begin{cases} x=m+1-\dfrac{m^2}{m+1}\\ y=\dfrac{m}{m+1} \end{cases}\Leftrightarrow \begin{cases} x=\dfrac{m^2+2m+1-m^2}{m+1}=\dfrac{2m+1}{m+1}\\ y=\dfrac{m}{m+1} \end{cases}$

Với $m\neq -1$

Để $x,y\in \mathbb{Z}\Leftrightarrow \dfrac{2m+1}{m+1}$ và $\dfrac{m}{m+1}\in \mathbb{Z}$

Xét $\dfrac{2m+1}{m+1}=2-\dfrac{1}{m+1}\in \mathbb{Z}$

$\Leftrightarrow m+1\in Ư(1)=\left\{-1;1\right\}$

$\Leftrightarrow m\in \left\{-2;0\right\}(1)$

Xét $\dfrac{m}{m+1}=1-\dfrac{1}{m+1}\in \mathbb{Z}$

$\Leftrightarrow m+1\in Ư(1)=\left\{-1;1\right\}$

$\Leftrightarrow m\in \left\{-2;0\right\}(2)$

Từ $(1)(2)\Rightarrow m\in \left\{-2;0\right\}$

Vậy $m\in \left\{-2;0\right\}$