Question

[Tường Nguyễn Thế]

Tìm nghiệm nguyên của phương trình : \(2x^2+4x=19-3y^2\)

Answer 1

[Rimuru tempest]

\(2x^2+4x-19+3y^2=0\)

\(\Leftrightarrow2x^2+4x+2-21+3y^2=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)+3y^2=21\)

\(\Leftrightarrow2\left(x+1\right)^2+3y^2=21\)

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 ........................................

Answer 2

[tran nguyen bao quan]

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)}