a) \(\dfrac{2}{x-3}+\dfrac{x-5}{x-1}=1\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)+\left(x-5\right)\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}=1\)
\(\Leftrightarrow2\left(x-1\right)+\left(x-5\right)\left(x-3\right)=\left(x-3\right)\left(x-1\right)\)
\(\Leftrightarrow2x-2+x^2-8x+15-x^2+4x-3=0\)
\(\Leftrightarrow-2x+10=0\) \(\Leftrightarrow x=5\)
b) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\) (2)
Ta có \(x^2-1=\left(x-1\right)\left(x+1\right)\)
ĐKXĐ: \(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
(2) \(\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-16}{x^2-1}=0\)
mà \(x^2-1\ne0\) để phương trính có nghĩa
\(\Leftrightarrow\left(x+1\right)^2=\left(x-1\right)^2-16=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-16=0\)
\(\Leftrightarrow4x-16=0\) \(\Leftrightarrow x=4\)
