ĐKXĐ:\(x,y\ne0\)
Đặt \(\dfrac{1}{\sqrt{x}}=a,\dfrac{1}{\sqrt{y}}=b\)
\(Hệ\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\dfrac{5}{3}\\5a+3b=\dfrac{13}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=\dfrac{1}{3}\\\dfrac{1}{\sqrt{y}}=\dfrac{1}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=9\\y=36\left(tm\right)\end{matrix}\right.\)
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