3 + root 7 upon 3 minus root 7 minus 3 minus root 7 upon 3 + root 7
Answers
Step-by-step explanation:
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Answer:
\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=\sqrt{5}
3+
5
7+3
5
−
3−
5
7−3
5
=
5
Step-by-step explanation:
Given: \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}
3+
5
7+3
5
−
3−
5
7−3
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}}
3+
5
7+3
5
−
3−
5
7−3
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}}=
3+
5
7+3
5
×
3−
5
3−
5
−
3−
5
7−3
5
×
3+
5
3+
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})}=
(3+
5
)(3−
5
)
(7+3
5
)(3−
5
)
−
(3−
5
)(3+
5
)
(7−3
5
)(3+
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}=
(3)
2
−(
5
)
2
21+9
5
−7
5
−3(5)
−
(3)
2
−(
5
)
2
21+7
5
−9
5
−3(5)
=\frac{6+2\sqrt{5}}{9-5}-\frac{6-2\sqrt{5}}{9-5}=
9−5
6+2
5
−
9−5
6−2
5
=\frac{6+2\sqrt{5}-(6-2\sqrt{5})}{4}=
4
6+2
5
−(6−2
5
)
=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}=
4
6+2
5
−6+2
5
=\frac{4\sqrt{5}}{4}=
4
4
5
=\sqrt{5}=
5
Therefore, \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=\sqrt{5}
3+
5
7+3
5
−
3−
5
7−3
5
=
5
Answer: