ĐKXĐ: x>=2 và y<>-x
\(\left\{{}\begin{matrix}\dfrac{4}{x+y}-2\sqrt{9x-18}=14\\\dfrac{5-2x-2y}{x+y}-\sqrt{4x-8}=-\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x+y}-6\sqrt{x-2}=14\\\dfrac{5}{x+y}-2-2\sqrt{x-2}=-\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x+y}-3\sqrt{x-2}=7\\\dfrac{5}{x+y}-2\sqrt{x-2}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{10}{x+y}-15\sqrt{x-2}=35\\\dfrac{10}{x+y}-4\sqrt{x-2}=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-11\sqrt{x-2}=38\left(vn\right)\\\dfrac{10}{x+y}-15\sqrt{x-2}=35\end{matrix}\right.\)
Vậy: Hệ vô nghiệm
ĐKXĐ: \(x\ge2;x+y\ne0\)
\(\left\{{}\begin{matrix}\dfrac{4}{x+y}-6\sqrt{x-2}=14\\\dfrac{5-2\left(x+y\right)}{x+y}-2\sqrt{x-2}=-\dfrac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}-3\sqrt{x-2}=7\\\dfrac{5}{x+y}-2-2\sqrt{x-2}=-\dfrac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}-3\sqrt{x-2}=7\\\dfrac{5}{x+y}-2\sqrt{x-2}=-\dfrac{3}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\sqrt{x-2}=v\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u-3v=7\\5u-2v=-\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=-\dfrac{37}{22}\\v=-\dfrac{38}{11}< 0\left(ktm\right)\end{matrix}\right.\)
Vậy hệ vô nghiệm
