Question

Tìm các cặp số nguyên (x;y) thỏa mãn 2x²+3y²+4x=19

Answer 1

\(2x^2+3y^2+4x=19\)

\(2x^2+4x=19-3y^2\)

\(2\left(x+1\right)^2=3\left(7-y^2\right)\)

Vì \(2\left(x+1\right)^2⋮2\) nên \(3\left(7-y^2\right)⋮2\) hay \(7-y^2⋮2\Rightarrow y^2\) lẻ(1)

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow7-y^2\ge0\Rightarrow y^2\le7\)\(\Rightarrow y^2\in\left\{1;4\right\}\)(2)

Từ (1) và (2), ta suy ra: \(y^2=1\)\(\Rightarrow y\in\left\{-1;1\right\}\)

Ta có: \(2\left(x+1\right)^2=3\left(7-y^2\right)\)

\(2\left(x+1\right)^2=18\)

\(\left(x+1\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x+1=3\Rightarrow x=2\\x+1=-3\Rightarrow x=-4\end{matrix}\right.\)

Các cặp số nguyên \(\left(x;y\right)=\left\{\left(2;1\right);\left(2;-1\right);\left(-4;1\right);\left(-4;-1\right)\right\}\)