Đặt \(\dfrac{x}{x+1}=a;\dfrac{1}{y+4}=b\) . Hệ trở thành :
\(\left\{{}\begin{matrix}3a-2b=4\\2a-5b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6a-4b=8\\6a-15b=27\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11b=-19\\3a-2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{19}{11}\\a=\dfrac{2}{11}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{2}{11}\\\dfrac{1}{y+4}=-\dfrac{19}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{2}{9};-\dfrac{87}{19}\right)\)
Đặt \(\dfrac{x}{x+1}=a;\dfrac{1}{y+4}=b\) . Hệ phương trình trở thành :
\(\left\{{}\begin{matrix}3a-2b=4\\2a-5b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6a-4b=8\\6a-15b=27\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11b=-19\\3a-2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{19}{11}\\a=\dfrac{2}{11}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{2}{11}\\\dfrac{1}{y+4}=-\dfrac{19}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\)
Vậy ....
