Đáp án + giải thích bước giải :
`x^2y - x^2 + y=2`
`⇔ x^2 (y -1) + y - 1 = 1`
`⇔ (x^2 + 1) (y - 1) = 1`
`⇔` \(\left\{ \begin{array}{l}x^2+1\\y-1=1\end{array} \right.\) `∈ Ư (1) = {1;-1}`
`* x^2 + 1 ∈ {1;-1}`
`↔` \(\left\{ \begin{array}{l}x^2+1=1 ⇔ x = 0 (TM)\\x^2+1=-1⇔x =∅ \end{array} \right.\)
`* y - 1 ∈ {1; -1}`
`↔` \(\left\{ \begin{array}{l}y-1=1 ⇔ y = 2 (TM)\\y - 1=-1⇔y = 0 (KTM) \end{array} \right.\)
Vậy `(x;y) ∈ {0;2}`
