\(\frac { \sqrt { 5 } +\sqrt { 3 } }{ \sqrt { 5 } -\sqrt { 3 } } =\frac { \sqrt { 5 } +\sqrt { 3 } }{ \sqrt { 5 } -\sqrt { 3 } } \times \frac { \sqrt { 5 } +\sqrt { 3 } }{ \sqrt { 5 } +\sqrt { 3 } } \)
\(=\frac { 8+2\sqrt { 15 } }{ 2 } =4+\sqrt { 15 } \)
\(a+b\sqrt { 15 } =4+\sqrt { 15 } \)
a = 4, b = 1

\(\frac { \sqrt { 5 } +\sqrt { 2 } }{ \sqrt { 5 } -\sqrt { 2 } } =\frac { \left( \sqrt { 5 } +\sqrt { 2 } \right) }{ \left( \sqrt { 5 } -\sqrt { 2 } \right) } \times \frac { \left( \sqrt { 5 } +\sqrt { 2 } \right) }{ \left( \sqrt { 5 } +\sqrt { 2 } \right) } \)
\(=\frac { { \left( \sqrt { 5 } \right) }^{ 2 }+{ \left( \sqrt { 2 } \right) }^{ 2 }+2\times \sqrt { 5 } \times \sqrt { 2 } }{ { \left( \sqrt { 5 } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 } } \)
\(=\frac { 5+2+2\sqrt { 10 } }{ 5-2 } \)
\(=\frac { 7+2\times 3.162 }{ 3 } \)
\(=\frac { 7+6.324 }{ 3 } =\frac { 13.324 }{ 3 } \)
= 4.441 (approx)

\(\frac { 1 }{ 1+\sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } +\sqrt { 3 } } +\frac { 2 }{ \sqrt { 3 } +\sqrt { 5 } } \)
\(=\frac { 1 }{ \left( \sqrt { 2 } +1 \right) } \times \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } +\frac { 1 }{ \left( \sqrt { 3 } +\sqrt { 2 } \right) } \times \frac { \left( \sqrt { 3 } -\sqrt { 2 } \right) }{ \left( \sqrt { 3 } -\sqrt { 2 } \right) } +\frac { 2 }{ \left( \sqrt { 5 } +\sqrt { 3 } \right) } \times \frac { \left( \sqrt { 5 } -\sqrt { 3 } \right) }{ \left( \sqrt { 5 } -\sqrt { 3 } \right) } \)
\(\frac { \left( \sqrt { 2 } -1 \right) }{ 2-1 } +\frac { \left( \sqrt { 3 } -\sqrt { 2 } \right) }{ 3-2 } +\frac { 2\left( \sqrt { 5 } -\sqrt { 3 } \right) }{ 5-3 } \)
= √2-1 + √3 - √2 + √5-√3 = √5 - 1

\(\sqrt{5+2\sqrt{6}}=\sqrt{3+2+2\sqrt{6}}\)
=\(\sqrt{(\sqrt{3}+\sqrt{2})^{2}}\)
=\(\sqrt{3}+\sqrt{2}\)
\(\sqrt{8-2\sqrt{15}}=\sqrt{5+3-2\sqrt{15}}\)
=\(\sqrt{(\sqrt{5}-\sqrt{3})^{2}}=\sqrt{5}-\sqrt{3}\)
\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}\)=\(\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}\)
=\(\sqrt{2}+\sqrt{5}\)

\({ \left( 729 \right) }^{ \frac { -1 }{ 6 } }={ \left( { 3 }^{ 6 } \right) }^{ \frac { -4 }{ 6 } }={ 3 }^{ -4 }\)
=\(\frac{1}{3}\)

\(\frac { 2\sqrt { 3 } +3-\sqrt { 21 } }{ 3 } \)

\(\frac { 14931 }{ 8910 } \)

\(\frac { 291 }{ 99 } \)

\((3+\sqrt { 3 } )(3-\sqrt { 3 } )={ \left( 3 \right) }^{ 2 }-{ \left( \sqrt { 3 } \right) }^{ 2 }\)
\(\\ =9-3=6\)

1(\(\neq 0\)) is a rational number and \(\sqrt { 2 } \)(\(\neq 0\)) is an irrational number.
\(\frac { 1 }{ \sqrt { 2 } } \) is an irrational number.
The quotient of a non-zero rational number with an irrational number is irrational.

Let x = \(0.\overline { 001 } \) = 0.001001001...
Multiplying both sides by 1000, we get
1000 x = 1.001001...
1000x = 1 + 0.001001001...
1000x = 1 + x
1000x - x = 1
999x = 1
x = 1/999
Thus, \(0.\overline { 001 } \)=1/999
Here p = 1
q = 999(\(\neq 0\))

\(\frac { 2 }{ 11 } \)= 0.1818...= \(0.\overline { 18 } \)

\(\frac { 3 }{ 13 } \)= 0.230769230769...=\(0.\overline { 230769 } \)

There can be infinitely many rational numbers between 3 and 4.
\(\frac { 3+4 }{ 2 } =\frac { 7 }{ 2 } \)
\(\\ \frac { 3+\frac { 7 }{ 2 } }{ 2 } =\frac { 13 }{ 4 } \)
\(\\ \frac { 3+\frac { 13 }{ 4 } }{ 2 } =\frac { 25 }{ 8 }\)
\( \\ \frac { 3+\frac { 25 }{ 8 } }{ 2 } =\frac { 49 }{ 16 } =\frac { 3+\frac { 49 }{ 16 } }{ 2 } =\frac { 97 }{ 32 } =\frac { 3+\frac { 97 }{ 32 } }{ 2 } =\frac { 193 }{ 64 } \)
Thus, six rational numbers between 3 and 4
\(\frac { 193 }{ 64 } ,\frac { 97 }{ 32 } ,\frac { 49 }{ 16 } ,\frac { 25 }{ 8 } ,\frac { 13 }{ 4 } \)and \(\frac { 7 }{ 2 } \)
Aliter
\(3=\frac { 3 }{ 1 } =\frac { 3\times 7 }{ 1\times 7 } =\frac { 21 }{ 7 } \)
\(\\ 4=\frac { 4 }{ 1 } =\frac { 4\times 7 }{ 1\times 7 } =\frac { 28 }{ 7 } \)
6 + 1 = 7
the six rational numbers between 3 and 4 can be taken as 
\(\frac { 22 }{ 7 } ,\frac { 23 }{ 7 } ,\frac { 24 }{ 7 } ,\frac { 25 }{ 7 } ,\frac { 26 }{ 7 } \) and \(\frac { 27 }{ 7 } \)

Yes! zero is a rational number.We can write zero in the form \(\frac { p }{ q } \),where p and q are integers and \(q\neq 0\)as follows:
  \(0=\frac { 0 }{ 1 } =\frac { 0 }{ 2 } =\frac { 0 }{ 3 } \)etc.