a: \(\left\{{}\begin{matrix}4\sqrt{5}x-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\sqrt{10}\cdot x-\sqrt{2}\cdot y=6\\10x+\sqrt{2}y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(4\sqrt{10}+10\right)=6-1=5\\4\sqrt{5}\cdot x-y=3\sqrt{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{10+4\sqrt{10}}=\dfrac{-5-2\sqrt{10}}{6}\\y=4\sqrt{5}\cdot x-3\sqrt{2}=\dfrac{-10\sqrt{5}+11\sqrt{2}}{3}\end{matrix}\right.\)
b: ĐKXĐ: y>2
\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\6\left|x-1\right|+\dfrac{3}{\sqrt{y-2}}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left|x-1\right|=14\\6\left|x-1\right|+\dfrac{3}{\sqrt{y-2}}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=2\\\dfrac{3}{\sqrt{y-2}}=15-6\cdot2=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1\in\left\{2;-2\right\}\\y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{3;1\right\}\\y=3\end{matrix}\right.\left(nhận\right)\)
