a, đk : x khác 1 ; -1
b, \(C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2\left(1-x^2\right)}=\dfrac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\)
c, Ta có C = 1/2
khi \(\dfrac{1}{2\left(x+1\right)}=\dfrac{1}{2}\Rightarrow x+1=1\Leftrightarrow x=0\)


