Simplify the expression

Factor the numerator, $\frac{(x-5)(x+5)}{x^2+10x+25}$ . Now factor the denominator, $\frac{(x-5)(x+5)}{(x+5)(x+5)}$ and finally cancel out like terms, $\frac{(x-5)}{(x+5)}$ $×$ $\frac{(x+5)}{(x+5)}$ $=$ $\frac{(x-5)}{(x+5)}$ $×(1)=$ $\frac{(x-5)}{(x+5)}$ .

We will use two formulas. $(a+b)^2 = a^2 + 2ab +b^2$ $a^2 - b^2 = (a+b)(a-b)$

By applying these, $\frac {x^{2} - 25}{x^{2} + 10x + 25} = \frac {x^{2} - 5^2}{x^2 + 2 \cdot 10x + 5^2 } = \frac {(x+5)(x-5)}{(x+5)^2} = \frac {x-5}{x+5}$

$\dfrac{x^2-25}{x^2+10x+25}=\dfrac{x^2-5^2}{(x+5)(x+5)}=\dfrac{(x+5)(x-5)}{(x+5)(x+5)}=$ $\color{#D61F06}\boxed{\dfrac{x-5}{x+5}}$

$\dfrac{x^2 - 25}{x^2 + 10x + 25} = \dfrac{(x + 5)(x - 5)}{(x + 5)(x + 5)} = \color{#69047E}{\boxed{\dfrac{(x - 5)}{(x + 5)}}}$ .