\(k,\) Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\y-2=b\ne0\end{matrix}\right.\), hpt trở thành:
\(\left\{{}\begin{matrix}3a-\dfrac{4}{b}=1\\a+\dfrac{2}{b}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-\dfrac{4}{b}=1\\2a+\dfrac{4}{b}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=5\\a+\dfrac{2}{b}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)
Từ đó thế vào tìm đc x,y
\(m,\) Đặt \(\left\{{}\begin{matrix}\left|x-1\right|=a\ge0\\y-2=b\ne0\end{matrix}\right.\), hpt trở thành:
\(\left\{{}\begin{matrix}2a+\dfrac{3}{b}=3\\3a-\dfrac{1}{b}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+\dfrac{3}{b}=3\\9a-\dfrac{3}{b}=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11a=33\\2a+\dfrac{3}{b}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)
Từ đó thế vào tìm x,y
