a) $\left\{\begin{matrix}\frac{1}{x} - \frac{1}{y} = \frac{1}{2}\\ \frac{3}{x} + \frac{4}{y} = 5\end{matrix}\right.$

Đặt $\frac{1}{x} = u;\;\frac{1}{y} = v \Rightarrow \left\{\begin{matrix}u - v = \frac{1}{2}\\ 3u + 4v = 5\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}4u - 4v = 2\\ 3u + 4v = 5\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}7u = 7\\ 3u + 4v = 5\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}u = 1\\ v = \frac{1}{2}\end{matrix}\right.$

$\Rightarrow \left\{\begin{matrix}x = 1\\ y = 2\end{matrix}\right.$

b) $\left\{\begin{matrix}\frac{1}{x + y} + \frac{1}{x - y} = \frac{5}{8}\\ \frac{1}{x - y} - \frac{1}{x + y} = \frac{3}{8}\end{matrix}\right.$

Đặt $\frac{1}{x + y} = u;\;\frac{1}{x - y} = v \Rightarrow \left\{\begin{matrix}u + v = \frac{5}{8}\\ u - v = \frac{3}{8}\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}2u = 1\\ u - v = \frac{3}{8}\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}u = \frac{1}{2}\\ v = \frac{1}{8}\end{matrix}\right.$

$\Rightarrow \left\{\begin{matrix}x + y = 2\\ x - y = 8\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}2x = 10\\ x - y = 8\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}x = 5\\ y = -3\end{matrix}\right.$

c) $\left\{\begin{matrix}\sqrt{x} + \sqrt{y - 1} = 3\\ 3\sqrt{x} - 4\sqrt{y - 1} = -5\end{matrix}\right.$

Đặt $\sqrt{x} = u;\;\sqrt{y - 1} = v \Rightarrow  \left\{\begin{matrix}u + v = 3\\ 3u - 4v = -5\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}4u + 4v = 12\\ 3u - 4v = -5\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}7u = 7\\ 3u - 4v = -5\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}u = 1\\ 3u - 4v = -5\end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}u = 1\\ v = 2\end{matrix}\right.$

$\Rightarrow  \left\{\begin{matrix}\sqrt{x} = 1\\ \sqrt{y - 1} = 2\end{matrix}\right.$

$\Leftrightarrow  \left\{\begin{matrix}x = 1\\ y -1 = 2\end{matrix}\right.$

$\Leftrightarrow  \left\{\begin{matrix}x = 1\\ y = 3\end{matrix}\right. $