Question

Answer 1

a: 2x-1=0

=>2x=1

=>\(x=\dfrac{1}{2}\)

b: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

\(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\)

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

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

=>\(x^2+2x+1-x+1=x^2+2\)

=>x+2=2

=>x=0(nhận)