2. Let $\log _{x}\left(2 x^{2}+x-1\right)>\log _{x} 2-1$. Then the range of $x$ is $(\quad)$.
(A) $\frac{1}{2}\frac{1}{2}, x \neq 1$
(C) $x>1$
(D) $0<x<1$
2. B.

Since $\left\{\begin{array}{l}x>0, x \neq 1, \\ 2 x^{2}+x-1>0,\end{array}\right.$ therefore, $x>\frac{1}{2}, x \neq 1$.
From $\log _{x}\left(2 x^{2}+x-1\right)>\log _{x} 2-1$, we deduce
$$
\log _{x}\left(2 x^{3}+x^{2}-x\right)>\log _{x} 2 \text {, }
$$

further leading to
$$
\left\{\begin{array} { l } 
{ 0 < x < 1, \\
2 x^{3}+x^{2}-x>2 .
\end{array}\right.\right.
$$

Solving, we get $0 < x < 1$ or $x > 1$.
Therefore, the range of $x$ is $x>\frac{1}{2}$, and $x \neq 1$.