Question

Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\dfrac{4}{x+y}+\dfrac{1}{y-1}=5\\\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\end{matrix}\right.\)

Answer 1

Lời giải:
Đặt $\frac{1}{x+y}=a; \frac{1}{y-1}=b$ thì hpt trở thành:
\(\left\{\begin{matrix} 4a+b=5\\ a-2b=-1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 8a+2b=10\\ a-2b=-1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 8a+2b+a-2b=9\\ a-2b=-1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=1\\ b=1\end{matrix}\right.\)

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