Tìm xy nguyên thoả mã a,y(x-3)+3x-9=11 b,xy+5y-2y-7=0 27/09/2021 Bởi Nevaeh Tìm xy nguyên thoả mã a,y(x-3)+3x-9=11 b,xy+5y-2y-7=0
\(\begin{array}{l}a,y\left( {x – 3} \right) + 3x – 9 = 11\\ \Leftrightarrow y\left( {x – 3} \right) + 3\left( {x – 3} \right) = 11\\ \Leftrightarrow \left( {x – 3} \right)\left( {y + 3} \right) = 11\\ + )\,\left\{ \begin{array}{l}x – 3 = 11\\y + 3 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 14\\y = – 2\end{array} \right.\\ + )\,\left\{ \begin{array}{l}x – 3 = 1\\y + 3 = 11\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 4\\y = 8\end{array} \right.\\ + )\,\left\{ \begin{array}{l}x – 3 = – 1\\y + 3 = – 11\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = 2\\x = – 14\end{array} \right.\\ + )\left\{ \begin{array}{l}x – 3 = – 11\\y + 3 = – 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = – 8\\y = – 4\end{array} \right.\end{array}\) Câu b) xem lại đề bài bạn nhé Bình luận
\(\begin{array}{l}a,y\left( {x – 3} \right) + 3x – 9 = 11\\ \Leftrightarrow y\left( {x – 3} \right) + 3\left( {x – 3} \right) = 11\\ \Leftrightarrow \left( {x – 3} \right)\left( {y + 3} \right) = 11\\ + )\,\left\{ \begin{array}{l}x – 3 = 11\\y + 3 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 14\\y = – 2\end{array} \right.\\ + )\,\left\{ \begin{array}{l}x – 3 = 1\\y + 3 = 11\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 4\\y = 8\end{array} \right.\\ + )\,\left\{ \begin{array}{l}x – 3 = – 1\\y + 3 = – 11\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = 2\\x = – 14\end{array} \right.\\ + )\left\{ \begin{array}{l}x – 3 = – 11\\y + 3 = – 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = – 8\\y = – 4\end{array} \right.\end{array}\)
Câu b) xem lại đề bài bạn nhé