\(\left\{{}\begin{matrix}x+y+\sqrt{x^2-y^2}=12\left(1\right)\\y\sqrt{x^2-y^2}=12\left(2\right)\end{matrix}\right.\)
ĐK: \(x^2\ge y^2\),y\(\ge0\)
(1)\(\Leftrightarrow y+\sqrt{x^2-y^2}=12-x\Leftrightarrow y^2+x^2-y^2+2y\sqrt{x^2-y^2}=144-24x+x^2\Leftrightarrow y\sqrt{x^2-y^2}=72-12x\)
Thế vào (2)\(\Leftrightarrow72-12x=12\Leftrightarrow12x=60\Leftrightarrow x=5\)
Vậy \(y\sqrt{25-y^2}=12\Leftrightarrow y^2\left(25-y^2\right)=144\Leftrightarrow y^4-25y^2+144=0\Leftrightarrow y^4-9y^2-16y^2+144=0\Leftrightarrow y^2\left(y^2-9\right)-16\left(y^2-9\right)=0\Leftrightarrow\left(y^2-9\right)\left(y^2-16\right)=0\Leftrightarrow\left(y-3\right)\left(y+3\right)\left(y-4\right)\left(y-4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y-3=0\\y+3=0\\y-4=0\\y+4=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}y=3\left(tm\right)\\y=-3\left(ktm\right)\\y=4\left(tm\right)\\y=-4\left(ktm\right)\end{matrix}\right.\)
Vậy (x;y)={(5;3);(5;4)}
