tìm x,y thuộc N sao cho : x+6=y(x-1)
2xy-x+4y=20
tìm x,y thuộc N sao cho : x+6=y(x-1) 2xy-x+4y=20
By aihong
By aihong
tìm x,y thuộc N sao cho : x+6=y(x-1)
2xy-x+4y=20
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
x + 6 = y\left( {x – 1} \right)\\
\Leftrightarrow y\left( {x – 1} \right) – \left( {x + 6} \right) = 0\\
\Leftrightarrow y\left( {x – 1} \right) – \left( {x – 1} \right) = 7\\
\Leftrightarrow \left( {x – 1} \right)\left( {y – 1} \right) = 7\\
7 = 1.7 = \left( { – 1} \right).\left( { – 7} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 1 = 1\\
y – 1 = 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 1 = – 1\\
y – 1 = – 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 1 = 7\\
y – 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 1 = – 7\\
y – 1 = – 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 2\\
y = 8
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 0\\
y = – 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 8\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = – 6\\
y = 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) \in \left\{ {\left( { – 6;0} \right);\left( {2;8} \right);\left( {0; – 6} \right);\left( {8;2} \right)} \right\}\\
b,\\
2xy – x + 4y = 20\\
\Leftrightarrow x\left( {2y – 1} \right) + 2.\left( {2y – 1} \right) = 20 – 2\\
\Leftrightarrow \left( {x + 2} \right)\left( {2y – 1} \right) = 18
\end{array}\)
\(y\) là số nguyên nên \(2y – 1\) là số lẻ.
Do đó, ta có:
\(\begin{array}{l}
\left( {x + 2} \right)\left( {2y – 1} \right) = 18\\
18 = 1.18 = 2.9 = 3.6 = \left( { – 1} \right).\left( { – 18} \right) = \left( { – 2} \right).\left( { – 9} \right) = \left( { – 3} \right).\left( { – 6} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2y – 1 = 1\\
x + 2 = 18
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = 9\\
x + 2 = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = 3\\
x + 2 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = – 1\\
x + 2 = – 18
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = – 3\\
x + 2 = – 6
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = – 9\\
x + 2 = – 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = 1\\
x = 16
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 5\\
x = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 2\\
x = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 0\\
x = – 20
\end{array} \right.\\
\left\{ \begin{array}{l}
y = – 1\\
x = – 8
\end{array} \right.\\
\left\{ \begin{array}{l}
y = – 4\\
x = – 4
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) \in \left\{ {\left( { – 4; – 4} \right);\left( { – 8; – 1} \right);\left( { – 20;0} \right);\left( {4;2} \right);\left( {0;5} \right);\left( {16;1} \right)} \right\}
\end{array}\)