rationalize the denominator 2 by root 5 + root 2 + 1 by root 3 + root 2 minus 3 by root 5 + root 2
Answers
Step-by-step explanation:
5
+
3
2
+
3
+
2
1
−
5
+
2
3
=0
Step-by-step explanation:
\begin{gathered}i) \frac{2}{\sqrt{5}+\sqrt{3}}\\=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}\\=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}\\= < /p > < p > \frac{2(\sqrt{5}-\sqrt{3})}{(5-3)}\\=\frac{2(\sqrt{5}-\sqrt{3})}{2}\\=\sqrt{5}-\sqrt{3}---(1)\end{gathered}
i)
5
+
3
2
=
(
5
+
3
)(
5
−
3
)
2(
5
−
3
)
=
(
5
)
2
−(
3
)
2
2(
5
−
3
)
=</p><p>
(5−3)
2(
5
−
3
)
=
2
2(
5
−
3
)
=
5
−
3
−−−(1)
\begin{gathered}ii) \frac{1}{\sqrt{3}+\sqrt{2}}\\=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\\= < /p > < p > \frac{(\sqrt{3}-\sqrt{2})}{(3-2)}\\=\frac{(\sqrt{3}-\sqrt{2})}{1}\\=\sqrt{3}-\sqrt{2}---(2)\end{gathered}
ii)
3
+
2
1
=
(
3
+
2
)(
3
−
2
)
(
3
−
2
)
=
(
3
)
2
−(
2
)
2
(
3
−
2
)
=</p><p>
(3−2)
(
3
−
2
)
=
1
(
3
−
2
)
=
3
−
2
−−−(2)
\begin{gathered}iii) \frac{3}{\sqrt{5}+\sqrt{2}}\\=\frac{3(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}\\=\frac{3(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}\\= < /p > < p > \frac{3(\sqrt{5}-\sqrt{2})}{(5-2)}\\=\frac{3(\sqrt{5}-\sqrt{2})}{3}\\=\sqrt{5}-\sqrt{2}---(3)\end{gathered}
iii)
5
+
2
3
=
(
5
+
2
)(
5
−
2
)
3(
5
−
2
)
=
(
5
)
2
−(
2
)
2
3(
5
−
2
)
=</p><p>
(5−2)
3(
5
−
2
)
=
3
3(
5
−
2
)
=
5
−
2
−−−(3)
\begin{gathered}Now,\\\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}\\=\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{2}-(\sqrt{5}-\sqrt{2})\\ [from \\(1),(2)\:and\:(3)]\end{gathered}
Now,
5
+
3
2
+
3
+
2
1
−
5
+
2
3
=
5
−
3
+
3
−
2
−(
5
−
2
)
[from
(1),(2)and(3)]
\begin{gathered}=\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{2}\\=0\end{gathered}
=
5
−
3
+
3
−
2
−
5
+
2
=0
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