Question

Tìm x,y thỏa mãn \(\left\{{}\begin{matrix}\left(x+\sqrt{2015+x^2}\right)\left(y+\sqrt{2015+x^2}\right)=2015\\3x^2+8y^2-12xy=23\end{matrix}\right.\)

Answer 1

Sửa \(y+\sqrt{2015+x^2}\rightarrow y+\sqrt{2015+y^2}\)

Ta có: \(\left(x+\sqrt{2015+x^2}\right)\left(y+\sqrt{2015+y^2}\right)=2015\)

\(\Leftrightarrow\left(x+\sqrt{2015+x^2}\right)\left(\sqrt{2015+x^2}-x\right)\left(y+\sqrt{2015+y^2}\right)=2015\left(\sqrt{2015+x^2}-x\right)\)

\(\Leftrightarrow2015\left(y+\sqrt{2015+y^2}\right)=2015\left(\sqrt{2015+x^2}-x\right)\)

\(\Leftrightarrow x+y=\sqrt{2015+x^2}-\sqrt{2015+y^2}\)

Tương tự ta cũng có: \(x+y=\sqrt{2015+y^2}-\sqrt{2015+x^2}\)

Cộng theo vế 2 đẳng thức trên ta có:

\(2\left(x+y\right)=0\Leftrightarrow x=-y\)

Thay \(x=-y\) vào \(pt\left(2\right)\) ta có:

\(23y^2=23\Leftrightarrow y=\pm1\Leftrightarrow x=\mp1\)