ĐKXĐ: x>=0 và x<>1
\(\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x}-1}+2\left|2y-1\right|=5\\\dfrac{2}{\sqrt{x}-1}-\left|2y-1\right|=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x}-1}+2\left|2y-1\right|=5\\\dfrac{4}{\sqrt{x}-1}-2\left|2y-1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-1}=7\\\dfrac{2}{\sqrt{x}-1}-\left|2y-1\right|=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}-1=1\\\left|2y-1\right|=2-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\2y-1\in\left\{1;-1\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\y\in\left\{1;0\right\}\end{matrix}\right.\)
ĐKXĐ: \(x\ge0;x\ne1\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}-1}=u\\\left|2y-1\right|=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3u+2v=5\\2u-v=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+2v=5\\4u-2v=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7u=7\\v=2u-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}-1}=1\\\left|2y-1\right|=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1=1\\\left[{}\begin{matrix}2y-1=1\\2y-1=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}2y=0\\2y=2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\\left[{}\begin{matrix}y=0\\y=1\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(4;0\right);\left(4;1\right)\)
