Question

Giai He phuong trinh:

\(\left\{{}\begin{matrix}\dfrac{5\left(x-1\right)}{x+2y}+\dfrac{3\left(y+1\right)}{x-2y}=8\\\dfrac{20\left(x-1\right)}{x+2y}-\dfrac{7\left(y+1\right)}{x-2y}=-6\end{matrix}\right.\)

Nguyễn Việt Lâm, Hạnh Hạnh, ...

Answer 1

Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2y}=a\\\dfrac{y+1}{x-2y}=b\end{matrix}\right.\) hệ trở thành:

\(\left\{{}\begin{matrix}5a+3b=8\\20a-7b=-6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{2}{5}\\b=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{x+2y}=\dfrac{2}{5}\\\dfrac{y+1}{x-2y}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5\left(x-1\right)=2\left(x+2y\right)\\y+1=2\left(x-2y\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=5\\2x-5y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{23}{11}\\y=\dfrac{7}{11}\end{matrix}\right.\)