Question

7+3
$< 7 + 3 \sqrt{5} > \div < 3 + \sqrt{5} > - < 7 - 3 \sqrt{5} > \div < \ 3 - \sqrt{5} >$
​

Answer 1

Answer:

√5

Step-by-step explanation:

hope it helps you ! mark as brainliest

Answer 2

Answer:

$\large{\red{\boxed{\sf \frac{7 + 3 \sqrt{5} }{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = \sqrt{5}}}}$

Step-by-step explanation:

Given that ;

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

Let a = $\sf \dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}}$ and,

b = $\sf \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}$

We need to find, (a - b) -

Consider a;

$\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } \\ \\ \implies \frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } \times \frac{3 - \sqrt{5} }{3 - \sqrt{5} } \\ \\ \implies \frac{(7 + 3 \sqrt{5} )(3 - \sqrt{5} )}{(3) {}^{2} - ( \sqrt{5}) {}^{2} } \\ \\ \implies \frac{35 + 9 \sqrt{5} - 7 \sqrt{5} - 15 }{9 - 5} \\ \\ \implies \frac{20 + 2 \sqrt{5} }{4} \\ \\ \implies \frac{2(10 + \sqrt{5}) }{4}\\ \\ \implies \frac{10 + \sqrt{5} }{2}$

Consider b;

$\frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } \\ \\ \implies \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } \\ \\ \implies \frac{(7 - 3 \sqrt{5})(3 + \sqrt{5})}{(3) {}^{2} - ( \sqrt{5}) {}^{2} } \\ \\ \implies \frac{35 + 7 \sqrt{5} - 9 \sqrt{5} - 15}{9 - 5} \\ \\ \implies \frac{20 - 2 \sqrt{5} }{4} \\ \\ \implies \frac{2(10 - \sqrt{5}) }{4} \\ \\ \implies \frac{10 - \sqrt{5} }{2}$

Now, find (a - b) -

$\implies \frac{10 + \sqrt{5} }{2} - \frac{10 - \sqrt{5} }{2} \\ \\ \implies \frac{10 + \sqrt{5} - (10 - \sqrt{5}) }{2} \\ \\ \implies \frac{10 + \sqrt{5} - 10 + \sqrt{5} }{2} \\ \\ \implies \frac{2 \sqrt{5}}{ 2} \\ \\ \boxed{ \bf \therefore \: \frac{7 + 3 \sqrt{5} }{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = \sqrt{5} }$