Giải câu 2 trang 22 sách toán VNEN lớp 9 tập 2.
a) $\left\{\begin{matrix}2(x - y) - 3(x + y) = 5\\ 3(x - y) +5(x + y) = -2\end{matrix}\right.$
Đặt $\left\{\begin{matrix}u = (x - y)\\ v = (x + y)\end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix}2u - 3v = 5\\ 3u +5v = -2\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}u = 1\\ v = -1\end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix}2u - 3v = 5\\ 3u + 5v = -2\end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix}x = 0\\ y = -1\end{matrix}\right.$
b) $\left\{\begin{matrix}\frac{2}{x - 2} + \frac{2}{y - 1} = 2\\ \frac{2}{x - 2} - \frac{3}{y - 1} = 1\end{matrix}\right.$
Đặt $\left\{\begin{matrix}u = \frac{1}{x - 2}\\ v = \frac{1}{y - 1}\end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix}2u + 2v = 2\\ 2u - 3y = 1\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}u = \frac{4}{5}\\ v = \frac{1}{4}\end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix}x = \frac{13}{4}\\ y = 6\end{matrix}\right.$
c) $\left\{\begin{matrix}x + y = 24\\ \frac{x}{9} + \frac{y}{27} = 2\frac{8}{9}\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x + y = 24\\ 3x + y = 78\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x = 24 - y\\ 3(24 - y) + y = 78\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x = 24 - y\\ 2y = -6\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x = 27\\ y = -3\end{matrix}\right.$
d) $\left\{\begin{matrix}\sqrt{x - 1} - 3\sqrt{y + 2} = 2\\ 2\sqrt{x - 1} + 5\sqrt{y + 2} = 15\end{matrix}\right.$
Đặt $\left\{\begin{matrix}u = \sqrt{x - 1}\\ v = \sqrt{y + 2}\end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix}u - 3v = 2\\ 2u + 5v = 15\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}u = 5\\ v = 1\end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix}x = 26\\ y = -1\end{matrix}\right.$