First, $\mathbb{Q}[\sqrt{2}] = \mathbb{Q}(\sqrt{2})$ since $\frac{1}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$.
and as for more complicated examples:
Let $\alpha = \sqrt{\sqrt{3}+\sqrt{2}}$, $A = \mathbb{Q}[\alpha]$ and $B = \mathbb{Q}(\alpha)$.
Then $\frac{1}{\alpha} = \sqrt{\sqrt{3}- \sqrt{2}} = 10\alpha^3- \alpha^7$, which shows that $A = B$.
My conjecture is that $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha)$ is always true as long as $\alpha$ is algebraic, but I do not have a proof of this claim, just trials of many difference cases, and I am wondering if someone has a counterexample? or maybe a known criterion for when this is true?
Assuming $\alpha\neq 0$ is algebraic, so $\mathbb{Q}[\alpha]$ is a finite-dimensional $\mathbb{Q}$-vectorspace. Multiplication by any $q(\alpha)=q_0+q_1\alpha+\dots+q_k\alpha^k\neq 0$ gives a $\mathbb{Q}$-linear endomorphism of $\mathbb{Q}[\alpha]$ which is injective, hence also surjective by finite-dimensionality. So $\frac{1}{q(\alpha)}\in\mathbb{Q}[\alpha]$ and $\mathbb{Q}[\alpha]=\mathbb{Q}(\alpha)$.