Bài 2:
a: Đặt \(\dfrac{4}{\sqrt{x}-3}=a\); \(\dfrac{1}{\left|2y-1\right|}=b\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+b=5\\a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-3=2\\\left|2y-1\right|=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=25\\y\in\left\{1;0\right\}\end{matrix}\right.\)
b: Đặt x+y=a; \(\sqrt{x+2}=b\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+b=7\\5a-2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+2b=14\\5a-2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=7-2a=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(7;-5\right)\)
