Toán Lớp 8: tìm x,y nguyên thỏa mãn đẳng thức:
a, xy+3x-2y-9=0
b,xy-x+3y-4 =0
giúp mk vs
huhu
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Toán Lớp 8: tìm x,y nguyên thỏa mãn đẳng thức:
a, xy+3x-2y-9=0
b,xy-x+3y-4 =0
giúp mk vs
huhu
Comments ( 1 )
a)xy + 3x – 2y – 9 = 0\\
\Leftrightarrow x.\left( {y + 3} \right) – 2y – 6 – 3 = 0\\
\Leftrightarrow x.\left( {y + 3} \right) – 2.\left( {y + 3} \right) = 3\\
\Leftrightarrow \left( {y + 3} \right)\left( {x – 2} \right) = 3 = 1.3 = \left( { – 1} \right).\left( { – 3} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y + 3 = 1\\
x – 2 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = 3\\
x – 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = – 1\\
x – 2 = – 3
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = – 3\\
x – 2 = – 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = – 2;x = 5\\
y = 0;x = 3\\
y = – 4;x = – 1\\
y = – 6;x = 1
\end{array} \right.\\
Vậy\,\left( {y;x} \right) = \left\{ {\left( { – 2;5} \right);\left( {0;3} \right);\left( { – 4; – 1} \right);\left( { – 6;1} \right)} \right\}\\
b)xy – x + 3y – 4 = 0\\
\Leftrightarrow x\left( {y – 1} \right) + 3y – 3 – 1 = 0\\
\Leftrightarrow x\left( {y – 1} \right) + 3\left( {y – 1} \right) = 1\\
\Leftrightarrow \left( {y – 1} \right)\left( {x + 3} \right) = 1 = 1.1 = \left( { – 1} \right).\left( { – 1} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y – 1 = 1\\
x + 3 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y – 1 = – 1\\
x + 3 = – 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 2;x = – 2\\
y = 0;x = – 4
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( { – 2;2} \right);\left( { – 4;0} \right)} \right\}
\end{array}$