Toán Lớp 8: Tìm các cặp số nguyên (x;y) thỏa mãn: x^2 + xy – 2016x -2017y -2018 = 0
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Toán Lớp 8: Tìm các cặp số nguyên (x;y) thỏa mãn: x^2 + xy – 2016x -2017y -2018 = 0
Comments ( 2 )
Ta có : x2+xy−2016x−2017y−2018=0
⇔x2+xy+x−1−2017x−2017y−2017=0
⇔x(x+y+1)−2017(x+y+1)=1
⇔(x−2017)(x+y+1)=1
⇒{x−2017=1 ; x+y+1=1
{x−2017=−1 ; x+y+1=−1
⇒[{x=2018 ; y=−2018
{x=2016 ; y=−2018
Vậy
Từ pt tương đương :
(x^2+xy+x) – (2017x + 2017y + 2017)=1
<=>x(x+y+1)-2017 (x+y+1) = 1
<=>(x+y+1)(x-2017)=1.1=(-1).(-1)
TH1 :
x+y+1=1, x-2017=1
<=>x+y=0, x=2018
<=>x=2018, y=-2018
TH2 :
x+y+1=-1,x-2017=-1
<=>x+y=-2, x=2016
<=>x=2016, y=-2018
Vậy (x;y)=(2018, -2018); (2016, -2018)