Question

\(\dfrac{7}{x-1}-\dfrac{1}{3\left(x+1\right)}=\dfrac{x}{x^2-1}\)

Answer 1

\(\dfrac{7}{x-1}-\dfrac{1}{3\left(x+1\right)}=\dfrac{x}{x^2-1}\)      

\(\Leftrightarrow\) \(\dfrac{7}{x-1}-\dfrac{1}{3\left(x+1\right)}=\dfrac{x}{\left(x-1\right)\left(x-1\right)}\)            ĐK : \(\text{x ≠ ± 1}\)

\(\Rightarrow21\left(x+1\right)-x+1=3x\) 

\(\Leftrightarrow\) \(21x+21-x+1=3x\)

\(\Leftrightarrow\) \(20x+22=3x\)

\(\Leftrightarrow\) \(20x-3=-22\)

\(\Leftrightarrow17x=-22\)

\(\Leftrightarrow x=-\dfrac{22}{17}\)

Vậy \(S=\) {\(-\dfrac{22}{17}\)}