Tìm các cặp số nguyên (x; y) thoả mãn : x^2 +xy-2016x-2017y-2018=0. 07/08/2021 Bởi Faith Tìm các cặp số nguyên (x; y) thoả mãn : x^2 +xy-2016x-2017y-2018=0.
Giải thích các bước giải: `x^2 +xy-2016x-2017y-2018=0` `<=>x^2+xy+x-2017x-2017y-2017-1=0` `<=>x(x+y+1)-2017(x+y+1)=1` `<=>(x-2017)(x+y+1)=1 (1)` Vì x; y là số nguyên nên (1) `<=>`\(\left[ \begin{array}{l}\left \{ {{x-2017=1} \atop {x+y+1=1}} \right. \\\left \{ {{x-2017=-1} \atop {x+y+1=-1}} \right.\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}\left \{ {{x=2018} \atop {y=-2018}} \right. \\\left \{ {{x=2016} \atop {y=-2018}} \right.\end{array} \right.\) Vậy `(x ; y) ∈ {(2018; -2018);(2016; -2018)}` Bình luận
Ta có : `x²+xy-2016x-2017y-2018=0` `⇒ x²+xy+x-1-2017x-2017y-2017=0` `⇒(x-2017)(x+y+1)=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 … $@Race2k53$ Xin hay nhất ! Bình luận
Giải thích các bước giải:
`x^2 +xy-2016x-2017y-2018=0`
`<=>x^2+xy+x-2017x-2017y-2017-1=0`
`<=>x(x+y+1)-2017(x+y+1)=1`
`<=>(x-2017)(x+y+1)=1 (1)`
Vì x; y là số nguyên nên (1) `<=>`\(\left[ \begin{array}{l}\left \{ {{x-2017=1} \atop {x+y+1=1}} \right. \\\left \{ {{x-2017=-1} \atop {x+y+1=-1}} \right.\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}\left \{ {{x=2018} \atop {y=-2018}} \right. \\\left \{ {{x=2016} \atop {y=-2018}} \right.\end{array} \right.\)
Vậy `(x ; y) ∈ {(2018; -2018);(2016; -2018)}`
Ta có :
`x²+xy-2016x-2017y-2018=0`
`⇒ x²+xy+x-1-2017x-2017y-2017=0`
`⇒(x-2017)(x+y+1)=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 …
$@Race2k53$
Xin hay nhất !