Question

Giải hệ phương trình

a,\(\dfrac{3}{x-1}+\dfrac{1}{y+2}=4 Và\dfrac{2}{x-1}-\dfrac{1}{y+2}=1\)

Answer 1

ĐKXĐ: x<>1 và y<>-2

\(\left\{{}\begin{matrix}\dfrac{3}{x-1}+\dfrac{1}{y+2}=4\\\dfrac{2}{x-1}-\dfrac{1}{y+2}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x-1}+\dfrac{1}{y+2}+\dfrac{2}{x-1}-\dfrac{1}{y+2}=4+1\\\dfrac{2}{x-1}-\dfrac{1}{y+2}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{x-1}=5\\\dfrac{1}{y+2}=\dfrac{2}{x-1}-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\\dfrac{1}{y+2}=\dfrac{2}{1}-1=1\end{matrix}\right.\)

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