Ta có phương trình ẩn y:
\(\frac{y+5}{y-1}-\frac{y+1}{y-3}=\frac{-8}{\left(y-1\right)\left(y-3\right)}\)\(\left(ĐK:y\ne1;y\ne3\right)\)
\(\Rightarrow\frac{\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)}{\left(y-1\right)\left(y-3\right)}=\frac{-8}{\left(y-1\right)\left(y-3\right)}\)
\(\Rightarrow\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)=-8\)
\(\Rightarrow\left(y^2+2y-15\right)-\left(y^2-1\right)=-8\)
\(\Rightarrow y^2+2y-15-y^2+1=-8\Leftrightarrow2y-14=-8\)
\(\Leftrightarrow2y=6\Leftrightarrow y=3\)(ktm)
Vậy không có y để \(\frac{y+5}{y-1}-\frac{y+1}{y-3}=\frac{-8}{\left(y-1\right)\left(y-3\right)}\)
\(\frac{y+5}{y-1}-\frac{y+1}{y-3}=\frac{-8}{\left(y-1\right)\left(y-3\right)}ĐKXĐ:y\ne1;3\)
\(\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)=-8\)
\(2y-14=-8\)
\(2y=6\)
\(y=3\)Theo ĐKXĐ => vô nghiệm
