Tìm tất cả các số nguyên (x; y) thỏa mãn x.(y – 1) + y = 2.
Giải thích
x(y – 1) + y = 2
⇔ x(y – 1) + (y – 1) = 2 – 1
⇔ (y – 1)(x + 1) = 1
⇔x+1=1y−1=1x+1=−1y−1=−1⇔x=0y=2x=−2y=0
Vậy (x; y) ∈ {(0; 2), (0; −2)}
x(y – 1) + y = 2
⇔ x(y – 1) + (y – 1) = 2 – 1
⇔ (y – 1)(x + 1) = 1
⇔x+1=1y−1=1x+1=−1y−1=−1⇔x=0y=2x=−2y=0
Vậy (x; y) ∈ {(0; 2), (0; −2)}