ĐKXĐ: x<>-1/2 và y<>1
\(\left\{{}\begin{matrix}\dfrac{y-1}{2x+1}+\dfrac{8x+4}{y-1}=4\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{\left(y-1\right)^2+4\left(2x+1\right)^2}{\left(2x+1\right)\left(y-1\right)}=4\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y^2-2y+1+16x^2+16x+4=4\left(2xy-2x+y-1\right)\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}16x^2+24x+y^2-6y-8xy+9=0\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(16x^2-8xy+y^2\right)+24x-6y+9=0\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(4x-y\right)^2+6\left(4x-y\right)+9=0\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(4x-y+3\right)^2=0\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-y=-3\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x=0\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\left(nhận\right)\)
