\(2x^2+4x-19+3y^2=0\)
\(\Leftrightarrow2x^2+4x+2-21+3y^2=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)+3y^2=21\)
\(\Leftrightarrow2\left(x+1\right)^2+3y^2=21\)
mà 21 phải = bội của 2 và 3
\(\Rightarrow2\left(x+1\right)^2+3y^2=2.3^2+3.1^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+1\right)^2=2.3^2\\3y^2=3.1^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=3^2\\y^2=1^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-3^2=0\\y^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+4\right)=0\\\left(y-1\right)\left(y+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\\\left[{}\begin{matrix}y=-1\\y=1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-1\end{matrix}\right.\end{matrix}\right.\)
vậy ........................................
2x2+4x=19-3y2⇔2x2+4x+2=21-3y2⇔2(x+1)2=3(7-y2)
Ta có 2(x+1)2⋮2⇒3(7-y2)⋮2⇒7-y2⋮2⇒y lẻ (1)
Ta lại có 2(x+1)2≥0⇒3(7-y2)≥0⇒7-y2≥0⇒y2≤7⇒y2∈{1;4} (2)
Từ (1),(2)⇒y2∈{1}⇒y∈{-1;1}
Ta có y2=1⇒2(x+1)2=3(7-y2)=18⇒(x+1)2=9⇒x+1=3 hoặc x+1=-3
⇒x=2 hoặc x=-4
Vậy {x,y}={(-1;2);(-1;-4);(1;2);(1;-4)}
