\(a,ĐK:x\ge2;y\ne-3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\ge0\\y+3=b\ne0\end{matrix}\right.\Leftrightarrow y=b-3\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}2a+\dfrac{b-3}{b}=1\\4a-\dfrac{3\left(b-3\right)}{b}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+\dfrac{2\left(b-3\right)}{b}=2\\4a-\dfrac{3\left(b-3\right)}{b}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3-b}{b}=9\\2a+\dfrac{b-3}{b}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2a-9=1\\b=\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=27\\y=\dfrac{3}{10}-3=-\dfrac{27}{10}\end{matrix}\right.\)
\(b,ĐK:x\ge-2;y\ne\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\2y-1=b\ne0\end{matrix}\right.\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{b}=5\\3a+\dfrac{2}{b}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+\dfrac{3}{b}=15\\3a+\dfrac{2}{b}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{b}=5\\\dfrac{1}{b}=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=2\\b=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=4\\2y-1=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tm\right)\\y=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
