Question

7 + 3 root 3 upon 3 + root 5 minus 7 minus 3 root 5 upon 3 minus root 5

Answer 1

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Answer 2

Answer:

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=\sqrt{5}$

Step-by-step explanation:

Given: $\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}$

We need to simplify the Given Expression.

We use rationalization of denominator to simply the given expression.

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}$

$=\frac{7+3\sqrt{5}}{3+\sqrt{5}}\times\frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\times\frac{3+\sqrt{5}}{3+\sqrt{5}}$

$=\frac{(7+3\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}-\frac{(7-3\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$

$=\frac{21+9\sqrt{5}-7\sqrt{5}-3(5)}{(3)^2-(\sqrt{5})^2}-\frac{21+7\sqrt{5}-9\sqrt{5}-3(5)}{(3)^2-(\sqrt{5})^2}$

$=\frac{6+2\sqrt{5}}{9-5}-\frac{6-2\sqrt{5}}{9-5}$

$=\frac{6+2\sqrt{5}-(6-2\sqrt{5})}{4}$

$=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}$

$=\frac{4\sqrt{5}}{4}$

$=\sqrt{5}$

Therefore, $\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=\sqrt{5}$