ĐKXĐ: \(\left\{{}\begin{matrix}x\ne3\\y\ne-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x-3}-\dfrac{4}{y+1}=5\\\dfrac{3}{x-3}+\dfrac{4}{y+1}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-3}+\dfrac{3}{x-3}=5-1=4\\\dfrac{1}{x-3}-\dfrac{4}{y+1}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x-3}=4\\\dfrac{4}{y+1}=\dfrac{1}{x-3}-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=1\\\dfrac{4}{y+1}=1-5=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-2\end{matrix}\right.\left(nhận\right)\)
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