ĐKXĐ: x<>-2 và y>2
\(\left\{{}\begin{matrix}\dfrac{3}{\left|x+2\right|}+\dfrac{1}{\sqrt{y-2}}=4\\\dfrac{2}{\left|x+2\right|}-\dfrac{1}{\sqrt{y-2}}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{\left|x+2\right|}+\dfrac{1}{\sqrt{y-2}}+\dfrac{2}{\left|x+2\right|}-\dfrac{1}{\sqrt{y-2}}=5\\\dfrac{2}{\left|x+2\right|}-\dfrac{1}{\sqrt{y-2}}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{\left|x+2\right|}=5\\\dfrac{1}{\sqrt{y-2}}=\dfrac{2}{\left|x+2\right|}-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|x+2\right|=1\\\dfrac{1}{\sqrt{y-2}}=2-1=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2\in\left\{1;-1\right\}\\y-2=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{-1;-3\right\}\\y=3\end{matrix}\right.\left(nhận\right)\)
\(Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{\left|x+2\right|}\left(x\ne-2\right)\\b=\dfrac{1}{\sqrt{y-2}}\left(y\ge2\right)\end{matrix}\right.\)
\(Có:\left\{{}\begin{matrix}\dfrac{3}{\left|x+2\right|}+\dfrac{1}{\sqrt{y-2}}=4\\\dfrac{2}{\left|x+2\right|}-\dfrac{1}{\sqrt{y-2}}=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}3a+b=4\\2a-b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5a=5\\2a-b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left|x+2\right|}=1\\\dfrac{1}{\sqrt{y-2}}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+2\right|=1\\\sqrt{y-2}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+2=1\\x+2=-1\end{matrix}\right.\\y-2=1^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-3\left(TM\right)\end{matrix}\right.\\y=3\left(TM\right)\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left\{\left(-1;3\right);\left(-3;3\right)\right\}\)
