Thank you for the answer. But you have answered not my question. Because in the Hilbert module $\ell_2(A)$ the norm is defined by $\lVert .\rVert=\lVert \sum {a_j}^*a_j \rVert^{\frac{1}{2}}$. And I knew today that $\ell_2(A)$ doesn't have Opial property: $$x_n=(0,0,\ldots , 0,\begin{bmatrix}0&0 \\ 0& 1\end{bmatrix} , 0 ,\ldots ) $$ (number of 0 before the matrix is n, so $x_n \to 0(weakly)$) and $$ y= (\begin{bmatrix} 1 & 0 \\ 0&0 \end{bmatrix} , 0 , 0 , \ldots ) $$ So $\liminf \lVert x_n\rVert = \liminf \lVert x_n-y\rVert$.

And $\ell_2(A)$ is not similar to $\ell^2(\mathbb{C})$. I don't know what I've to do? Should I prove it or give a counterexample?

In Hilbert spaces we use parallelogram property to show that a Hilbert space has Opial property . In the space $\ell^p$ we use uniformly convex property to show that they have Opial property. But in the Hilbert $A$-module $\ell^2(A)$ we don't have parallelogram and uniformly-convex properties.

The Opial property in the Hilbert A-module $\ell_2(A)$ is equivalent to $ \liminf \lVert \langle x_n,x_n\rangle \rVert < \liminf \lVert \langle x_n,x_n\rangle +\langle y,y \rangle \rVert $ for all $y\ne 0$.