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Answer:
Identity used :
(a + b)(a - b) = {a}^{2} - {b}^{2}(a+b)(a−b)=a
2
−b
2
\begin{gathered}\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } + \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + b\sqrt{5} \\ \\\end{gathered}
3+
5
7+3
5
+
3−
5
7−3
5
=a+b
5
L.H.S,
On rationalizing the denominator we get,
\begin{gathered}= \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)}^{2} - {( \sqrt{5}) }^{2} } + \frac{7(3 + \sqrt{5} ) - 3 \sqrt{5} (3 + \sqrt{5} )}{ {(3)}^{2} - {( \sqrt{5}) }^{2} } \\ \\ = \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15}{9 - 5} + \frac{21 + 7 \sqrt{5} - 9 \sqrt{5} - 15}{9 - 5} \\ \\ = \frac{6 + 2 \sqrt{5} }{4} + \frac{ 6 - 2 \sqrt{5} }{4} \\ \\ = \frac{3 + \sqrt{5} }{2} + \frac{3 - \sqrt{5} }{2} \\ \\ = \frac{3 + \sqrt{5} + 3 - \sqrt{5} }{2} \\ \\ = \frac{6}{2} \\ \\ = 3\end{gathered}
=
3+
5
7+3
5
×
3−
5
3−
5
+
3−
5
7−3
5
×
3+
5
3+
5
=
(3)
2
−(
5
)
2
7(3−
5
)+3
5
(3−
5
)
+
(3)
2
−(
5
)
2
7(3+
5
)−3
5
(3+
5
)
=
9−5
21−7
5
+9
5
−15
+
9−5
21+7
5
−9
5
−15
=
4
6+2
5
+
4
6−2
5
=
2
3+
5
+
2
3−
5
=
2
3+
5
+3−
5
=
2
6
=3
On comparing we get,
a = 3
b = 0
Step-by-step explanation:
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