Tìm tất cả các cặp số nguyên (x; y) thỏa mãn x ² + xy -2016x – 2017x – 2018 = 0 08/08/2021 Bởi Elliana Tìm tất cả các cặp số nguyên (x; y) thỏa mãn x ² + xy -2016x – 2017x – 2018 = 0
Giải thích các bước giải: `x^2+ xy -2016x – 2017y – 2018 = 0` `=>(x^2+ xy+1) -(2017x + 2017y + 2017) -1=0` `=>x(x+ y+1) -2017(x + y + 1) -1=0` `=>(x-2017)(x+ y+1)=1` `=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}x-2017=1\\x+y+1=1\end{matrix}\right.\\\left\{\begin{matrix}x-2017=-1\\x+y+1=-1\end{matrix}\right.\end{array} \right.\) `=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}x=2018\\y=-2018\end{matrix}\right.\\\left\{\begin{matrix}x=2016\\y=-2018\end{matrix}\right.\end{array} \right.\) Vậy `(x;y)in{(2018;-2018);(2016;-2018)}.` Bình luận
Đáp án: Giải thích các bước giải: Ta có : \(x^2+xy-2016x-2017y-2018=0\) \(\Leftrightarrow x^2+xy+x-1-2017x-2017y-2017=0\) \(\Leftrightarrow x\left(x+y+1\right)-2017\left(x+y+1\right)=1\) \(\Leftrightarrow\left(x-2017\right)\left(x+y+1\right)=1\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2017=1\\x+y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2017=-1\\x+y+1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2018\\y=-2018\end{matrix}\right.\\\left\{{}\begin{matrix}x=2016\\y=-2018\end{matrix}\right.\end{matrix}\right.\) Vậy \(\left(x,y\right)\in\left\{\left(2018,-2018\right),\left(2016,-2018\right)\right\}\) Bình luận
Giải thích các bước giải:
`x^2+ xy -2016x – 2017y – 2018 = 0`
`=>(x^2+ xy+1) -(2017x + 2017y + 2017) -1=0`
`=>x(x+ y+1) -2017(x + y + 1) -1=0`
`=>(x-2017)(x+ y+1)=1`
`=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}x-2017=1\\x+y+1=1\end{matrix}\right.\\\left\{\begin{matrix}x-2017=-1\\x+y+1=-1\end{matrix}\right.\end{array} \right.\) `=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}x=2018\\y=-2018\end{matrix}\right.\\\left\{\begin{matrix}x=2016\\y=-2018\end{matrix}\right.\end{array} \right.\)
Vậy `(x;y)in{(2018;-2018);(2016;-2018)}.`
Đáp án:
Giải thích các bước giải:
Ta có : \(x^2+xy-2016x-2017y-2018=0\)
\(\Leftrightarrow x^2+xy+x-1-2017x-2017y-2017=0\)
\(\Leftrightarrow x\left(x+y+1\right)-2017\left(x+y+1\right)=1\)
\(\Leftrightarrow\left(x-2017\right)\left(x+y+1\right)=1\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2017=1\\x+y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2017=-1\\x+y+1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2018\\y=-2018\end{matrix}\right.\\\left\{{}\begin{matrix}x=2016\\y=-2018\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(2018,-2018\right),\left(2016,-2018\right)\right\}\)