ĐKXĐ:\(x\ge1\)
\(\left\{{}\begin{matrix}\sqrt{x-1}+3y=5\\3\sqrt{x-1}-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+9y=15\\3\sqrt{x-1}-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}14y=14\\3\sqrt{x-1}-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\3\sqrt{x-1}-5.1=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\3\sqrt{x-1}=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=5\end{matrix}\right.\)
Đặt \(\sqrt{x-1}=a\)
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}a+3y=5\\3a-5y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)
\(ĐK:x\ge1\)
Đặt \(\sqrt{x-1}=a;a\ge0\)
hpt trở thành:
\(\left\{{}\begin{matrix}a+3y=5\\3a-5y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-3y\\3\left(5-3y\right)-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-3y\\15-9y-5y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-3y\\14y=14\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\left(tm\right)\\y=1\end{matrix}\right.\)
ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{4}\) \(\Leftrightarrow x-1=4\) \(\Leftrightarrow x=5\left(tm\right)\)
Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)
