Question

simplify. rationalize the denominator.\\(\frac{sqrt{5}}{-7 + sqrt{7}}\\)

Explanation:

Step1: Identify conjugate of denominator

The denominator is $-7 + \sqrt{7}$, its conjugate is $-7 - \sqrt{7}$. Multiply numerator and denominator by this conjugate.
$$\frac{\sqrt{5}}{-7 + \sqrt{7}} \times \frac{-7 - \sqrt{7}}{-7 - \sqrt{7}}$$

Step2: Multiply numerators

Multiply the numerators: $\sqrt{5} \times (-7 - \sqrt{7}) = -7\sqrt{5} - \sqrt{35}$

Step3: Multiply denominators (use difference of squares)

Denominator: $(-7)^2 - (\sqrt{7})^2 = 49 - 7 = 42$

Step4: Combine results

Put numerator over denominator: $\frac{-7\sqrt{5} - \sqrt{35}}{42}$ (or factor out negative sign: $\frac{ - (7\sqrt{5} + \sqrt{35}) }{42} = \frac{7\sqrt{5} + \sqrt{35}}{-42}$ but typically written with positive denominator by factoring out -1 from numerator and denominator)
Wait, actually, let's re - check the sign in step 3. Wait, $(-7 + \sqrt{7})(-7 - \sqrt{7}) = (-7)^2 - (\sqrt{7})^2 = 49 - 7 = 42$, that's correct. And numerator: $\sqrt{5}(-7 - \sqrt{7})=-7\sqrt{5}-\sqrt{35}$. So we can factor out a negative sign from the numerator: $- (7\sqrt{5}+\sqrt{35})$, so the fraction is $\frac{ - (7\sqrt{5}+\sqrt{35}) }{42}=\frac{7\sqrt{5}+\sqrt{35}}{-42}=-\frac{7\sqrt{5}+\sqrt{35}}{42}$. Alternatively, we can write it as $\frac{7\sqrt{5}+\sqrt{35}}{-42}$ but the standard form is with a positive denominator, so we can multiply numerator and denominator by - 1: $\frac{7\sqrt{5}+\sqrt{35}}{-42}=\frac{-(7\sqrt{5}+\sqrt{35})}{42}=-\frac{7\sqrt{5}+\sqrt{35}}{42}$ or we can distribute the negative sign in the numerator: $\frac{-7\sqrt{5}-\sqrt{35}}{42}$

Answer:

$\boxed{-\dfrac{7\sqrt{5} + \sqrt{35}}{42}}$ (or equivalent forms like $\dfrac{-7\sqrt{5}-\sqrt{35}}{42}$)