a) \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
\(ĐKXĐ:\)\(x\ne1\)và \(x\ne3\)
\(\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)9x-3}=\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\frac{8}{\left(x-3\right)\left(x-1\right)}\)
\(\Leftrightarrow\)\(x^2-3x+5x-15=x^2-x+x-1-8\)
\(\Leftrightarrow\)\(x^2-3x+5x-15-x^2+x-x+1+8=0\)
\(\Leftrightarrow\)\(2x-6=0\)
\(\Leftrightarrow\)\(2x=6\)
\(\Leftrightarrow\)\(x=3\)( loại )
Vậy \(S=\varnothing\)
b) \(\frac{y+1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)
\(ĐKXĐ:\)\(y\ne2\)và \(y\ne-2\)
\(\frac{\left(y+1\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}-\frac{5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12}{\left(y-2\right)\left(y+2\right)}+\frac{\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}\)
\(\Leftrightarrow\)\(y^2+2y+y+2-5y+10=12+y^2-4\)
\(\Leftrightarrow\)\(y^2+2y+y+2-5y+10-10-12-y^2+4=0\)
\(\Leftrightarrow\)\(-2y+4=0\)
\(\Leftrightarrow\)\(-2y=-4\)
\(\Leftrightarrow\)\(y=2\)( loại 0
Vậy \(S=\varnothing\)
