Let $R$ a commutative ring and $I$ and ideal of $R$,how i can show that if $\sqrt{I}=R$ then $I=R$ Is trivial show that $I \subseteq R$, but i'm stuck proving that $R\subseteq I$.
If i take a $x\in R \implies x\in \sqrt{I} \implies \exists n \in \mathbb{N}; x^n\in I$
But, how i can show that $x \in I$ ? 
 A: Consider the ring $N$ of upper triangular (nilptent) matrices $M(2,\mathbb{R})$ with $0$ at the diagonal. for every element of $M$ of $N$, $M^2=0$, so $\sqrt 0=N$, but $0$ is not $N$.
A: Assuming your rings must have a multiplicative identity...
Then your assumptions give that $1 \in R = \sqrt{I}$. Hence $1^n \in I$ for some $n \in \mathbb{N}$. But $1^n=1 \in I$, so $I=R$ by the usual argument.  
If your rings need not have identity then it is false by (a) rschwieb's clever counterexample or (b) Tsemo's if you strengthen the problem not to require commutativity (in both examples the rings conveniently did not have an identity).
rschwieb advocates $R=2\mathbb{Z}/8\mathbb{Z}$ as a good commutative counterexample.  This is clearly commutative, but the cube of every element is $0$.  An element takes the form $2x+8\mathbb{Z}$, and the cube is
$$
\left(2x+8\mathbb{Z}\right)^3=(2x)^3 + 8\mathbb{Z} = 8x^3 +8\mathbb{Z}=0+8\mathbb{Z}.
$$
Hence the ideal $I=\{0\}$ gives a counterexample to your original statement since its radical is $R=2\mathbb{Z}/8\mathbb{Z}$ but $I$ is not.
We should be able to give away rep points for assists...
