Question

\(\dfrac{X+1}{X-1}-\dfrac{X-1}{X+1}=\dfrac{4}{\left(X-1\right)\cdot\left(X+1\right)}\)

Answer 1

\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\)

\(\Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\)

\(\Leftrightarrow2.2x=4\)

\(\Leftrightarrow x=1\) (loại)

Vậy tập nghiệm của pt là :

\(S=\varnothing\)