Tìm bộ 3 số nguyên x,y,z thỏa mãn hệ phương trình :
x + yz = 2020
{
xy + z = 2021
0 bình luận về “Tìm bộ 3 số nguyên x,y,z thỏa mãn hệ phương trình :
x + yz = 2020
{
xy + z = 2021”
$\begin{array}{l} \left\{ \begin{array}{l} x + yz = 2020\\ xy + z = 2021 \end{array} \right.\\ \Rightarrow x + yz + 1 = xy + z \Leftrightarrow x – xy + yz – z = – 1\\ \Leftrightarrow x\left( {1 – y} \right) – z\left( {1 – y} \right) = – 1\\ \Leftrightarrow \left( {1 – y} \right)\left( {x – z} \right) = – 1\\ \Leftrightarrow \left\{ \begin{array}{l} 1 – y = – 1\\ x – z = 1 \end{array} \right.hoặc \left\{ \begin{array}{l} 1 – y = 1\\ x – z = – 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} y = 2\\ x = z + 1 \end{array} \right. hoặc\left\{ \begin{array}{l} y = 0\\ x = z – 1 \end{array} \right.\\ \end{array}$
Với $\left\{ \begin{array}{l} y = 2\\ x = z + 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x + 2z = 2020\\ 2x + z = 2021 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 674\\ y = 2\\ z = 673 \end{array} \right.(TM)$
Với $\begin{array}{l} \left\{ \begin{array}{l} y = 0\\ x = z – 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 2020\\ y = 0\\ z = 2021 \end{array} \right.\\ \end{array}$
$\begin{array}{l} \left\{ \begin{array}{l} x + yz = 2020\\ xy + z = 2021 \end{array} \right.\\ \Rightarrow x + yz + 1 = xy + z \Leftrightarrow x – xy + yz – z = – 1\\ \Leftrightarrow x\left( {1 – y} \right) – z\left( {1 – y} \right) = – 1\\ \Leftrightarrow \left( {1 – y} \right)\left( {x – z} \right) = – 1\\ \Leftrightarrow \left\{ \begin{array}{l} 1 – y = – 1\\ x – z = 1 \end{array} \right.hoặc \left\{ \begin{array}{l} 1 – y = 1\\ x – z = – 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} y = 2\\ x = z + 1 \end{array} \right. hoặc\left\{ \begin{array}{l} y = 0\\ x = z – 1 \end{array} \right.\\ \end{array}$
Với $\left\{ \begin{array}{l} y = 2\\ x = z + 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x + 2z = 2020\\ 2x + z = 2021 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 674\\ y = 2\\ z = 673 \end{array} \right.(TM)$
Với $\begin{array}{l} \left\{ \begin{array}{l} y = 0\\ x = z – 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 2020\\ y = 0\\ z = 2021 \end{array} \right.\\ \end{array}$
Vậy $(x;y;z)=(674;2;673),(2020;0;2021)$