Question

\(\left\{{}\begin{matrix}x+y+\sqrt{x^2-y^2}=12\\y\sqrt{x^2-y^2}=12\end{matrix}\right.\)

Answer 1

ĐKXĐ: ...

\(y\sqrt{x^2-y^2}=12>0\Rightarrow y>0\)

\(y+\sqrt{x^2-y^2}=12-x\left(x\le12\right)\)

\(\Leftrightarrow y^2+x^2-y^2+2y\sqrt{x^2-y^2}=x^2-24x+144\)

\(\Leftrightarrow y\sqrt{x^2-y^2}=-12x+72\)

\(\Rightarrow-12x+72=12\Rightarrow x=5\)

\(\Rightarrow y\sqrt{25-y^2}=12\Rightarrow y...\) (bình phương 2 vế giải pt trùng phương)

Answer 2

đặt \(\sqrt{x+y}=a,\sqrt{x-y}=b\) ta có hệ phương trình sau \(\left\{{}\begin{matrix}a^2+ab=12\\\left(a^2-b^2\right)ab=12\end{matrix}\right.\) \(\Leftrightarrow a^2+ab-ab\left(a^2-b^2\right)=0\) \(\Leftrightarrow a\left(a+b\right)-\left(a+b\right)\left(a-b\right)ab=0\) \(\Leftrightarrow\left(a+b\right)\left(a-a^2b+ab^2\right)=0\)