Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+5}\\b=\sqrt{y+5}\end{matrix}\right.\) (a,b>0)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b=12\left(1\right)\\b^2+a=12\left(2\right)\end{matrix}\right.\)
lấy trên trừ dưới \(\Leftrightarrow a^2+b-b^2-a=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(3\right)\\a+b=1\left(4\right)\end{matrix}\right.\)
từ (1)(3) ta có hệ: \(\left\{{}\begin{matrix}a=b\\a^2+b=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\a^2+a-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left[{}\begin{matrix}a=3\\b=-4\left(l\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+5}=3\\\sqrt{y+5}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
từ (1)(4) ta có hệ \(\left\{{}\begin{matrix}a+b=1\\a^2+b=12\end{matrix}\right.\)
giải tương tự nha bạn
