tìm số tự nhiên x và y biết x+6=y(x-1)(làm theo dạng a.b=c nha) 16/10/2021 Bởi Amara tìm số tự nhiên x và y biết x+6=y(x-1)(làm theo dạng a.b=c nha)
Đáp án: $\begin{array}{l}x + 6 = y.\left( {x – 1} \right)\\ \Rightarrow x – 1 + 7 = y.\left( {x – 1} \right)\\ \Rightarrow x – 1 – y.\left( {x – 1} \right) = – 7\\ \Rightarrow \left( {x – 1} \right).\left( {1 – y} \right) = – 7\\ \Rightarrow \left( {x – 1} \right).\left( {y – 1} \right) = 7 = 1.7\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x – 1 = 1\\y – 1 = 7\end{array} \right.\\\left\{ \begin{array}{l}x – 1 = 7\\y – 1 = 1\end{array} \right.\\\left\{ \begin{array}{l}x – 1 = – 1\\y – 1 = – 7\end{array} \right.\\\left\{ \begin{array}{l}x – 1 = – 7\\y – 1 = – 1\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 2\\y = 8\end{array} \right.\\\left\{ \begin{array}{l}x = 8\\y = 2\end{array} \right.\\\left\{ \begin{array}{l}x = 0\\y = – 6\end{array} \right.\\\left\{ \begin{array}{l}x = – 6\\y = 0\end{array} \right.\end{array} \right.\\Vậy\,\left( {x;y} \right) = \left\{ {\left( {2;8} \right);\left( {8;2} \right);\left( {0; – 6} \right);\left( { – 6;0} \right)} \right\}\end{array}$ Bình luận
`x+6=y(x-1)` `⇒x+6-y(x-1)=0` `⇒x-1-y(x-1)=-7` `⇒(x-1)(1-y)=-7` `⇒(x-1)(y-1)=7` `⇒x-1∈Ư(7)={±1,±7}` `+)x-1=-1⇒x=0` `⇒y-1=-7⇒y=-6` `+)x-1=1⇒x=2` `⇒y-1=7⇒y=8` `+)x-1=-7⇒x=-6` `⇒y-1=-1⇒y=0` `+)x-1=7⇒x=8` `⇒y-1=1⇒y=2` Vậy `(x,y)` là:`(0,-6);(2,8);(-6,0);(8,2)` Bình luận
Đáp án:
$\begin{array}{l}
x + 6 = y.\left( {x – 1} \right)\\
\Rightarrow x – 1 + 7 = y.\left( {x – 1} \right)\\
\Rightarrow x – 1 – y.\left( {x – 1} \right) = – 7\\
\Rightarrow \left( {x – 1} \right).\left( {1 – y} \right) = – 7\\
\Rightarrow \left( {x – 1} \right).\left( {y – 1} \right) = 7 = 1.7\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 1 = 1\\
y – 1 = 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 1 = 7\\
y – 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 1 = – 1\\
y – 1 = – 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 1 = – 7\\
y – 1 = – 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 2\\
y = 8
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 8\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 0\\
y = – 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = – 6\\
y = 0
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {2;8} \right);\left( {8;2} \right);\left( {0; – 6} \right);\left( { – 6;0} \right)} \right\}
\end{array}$
`x+6=y(x-1)`
`⇒x+6-y(x-1)=0`
`⇒x-1-y(x-1)=-7`
`⇒(x-1)(1-y)=-7`
`⇒(x-1)(y-1)=7`
`⇒x-1∈Ư(7)={±1,±7}`
`+)x-1=-1⇒x=0`
`⇒y-1=-7⇒y=-6`
`+)x-1=1⇒x=2`
`⇒y-1=7⇒y=8`
`+)x-1=-7⇒x=-6`
`⇒y-1=-1⇒y=0`
`+)x-1=7⇒x=8`
`⇒y-1=1⇒y=2`
Vậy `(x,y)` là:`(0,-6);(2,8);(-6,0);(8,2)`