Question

[Nguyễn Hoàng Minh]

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Answer 1

[Nguyễn Đức Trí]

\(2x^2+8xy+6y^2=24\left(x;y\inℕ^∗\right)\)

\(\Rightarrow2\left(x^2+4xy+3y^2\right)=24\)

\(\Rightarrow\left(x^2+4xy+4y^2\right)-y^2=12\)

\(\Rightarrow\left(x+2y\right)^2-y^2=12\)

\(\Rightarrow\left(x+2y+y\right)\left(x+2y-y\right)=12\)

\(\Rightarrow\left(x+3y\right)\left(x+y\right)=12\)

\(\Rightarrow\left(x+3y\right);\left(x+y\right)\in\left\{1;2;3;4;6;12\right\}\)

\(\Rightarrow\left(x+3y\right);\left(x+y\right)\in\left\{1;2;3;4;6;12\right\}\)

Ta giải các hệ phương trình sau :

1) \(\left\{{}\begin{matrix}x+3y=1\\x+y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=-11\left(loại\right)\\x+y=12\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}x+3y=2\\x+y=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=-3\left(loại\right)\\x+y=6\end{matrix}\right.\)

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

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

5) \(\left\{{}\begin{matrix}x+3y=6\\x+y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=4\\x+y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=2\\x=0\left(loại\right)\end{matrix}\right.\)

6) \(\left\{{}\begin{matrix}x+3y=12\\x+y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y=11\left(loại\right)\\x+y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\varnothing\right\}\)