Question

[Khaaaaaa]

Tìm x, y là số nguyên để 2(x+1)²+3y²=21

Answer 1

[Vô danh]

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

Ta có: 2(x+1)²+3y²=21≥3y²⇒y²≤7

Mà y² là SCP nên \(y^2\in\left\{0;1;4\right\}\)

Với \(y^2=0\Rightarrow y=0\), thay vào (1) ta có:

\(2\left(x+1\right)^2+3.0^2=21\\ \Rightarrow2\left(x+1\right)^2=21\\ \Rightarrow\left(x+1\right)^2=\dfrac{21}{2}\left(ktm\right)\)

Với \(y^2=1\Rightarrow y=\pm1\), thay vào (1) ta có:

\(2\left(x+1\right)^2+3.\left(\pm1\right)^2=21\\ \Rightarrow2\left(x+1\right)^2+3=21\\ \Rightarrow\left(x+1\right)^2=9\\ \Rightarrow\left[{}\begin{matrix}x+1=-3\\x+1=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Với \(y^2=4\Rightarrow y=\pm2\), thay vào (1) ta có:

\(2\left(x+1\right)^2+3.\left(\pm2\right)^2=21\\ \Rightarrow2\left(x+1\right)^2+12=21\\ \Rightarrow\left(x+1\right)^2=\dfrac{9}{2}\left(ktm\right)\)

Vậy \(\left(x,y\right)\in\left\{\left(-4;1\right);\left(-4;-1\right);\left(2;1\right);\left(2;-1\right)\right\}\)