ĐKXĐ: x<>2 và y<>-1
\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2x-2}{x-2}+\dfrac{y+2}{y+1}=\dfrac{26}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2x-4+2}{x-2}+\dfrac{y+1+1}{y+1}=\dfrac{26}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\2+\dfrac{2}{x-2}+1+\dfrac{1}{y+1}=\dfrac{26}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=\dfrac{11}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{4}{x-2}+\dfrac{2}{y+1}=\dfrac{22}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{1}{x-2}=-\dfrac{5}{5}=-1\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=\dfrac{11}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2=1\\\dfrac{1}{y+1}=\dfrac{11}{5}-2=\dfrac{1}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2=1\\y+1=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\left(nhận\right)\)
