\(\frac { \sqrt { 3 } +\sqrt { 2 } }{ 5+\sqrt { 2 } } =\frac { \sqrt { 3 } +\sqrt { 2 } }{ 5+\sqrt { 2 } } \times \frac { 5-\sqrt { 2 } }{ 5-\sqrt { 2 } } \)
\(=\frac { 5\sqrt { 3 } +5\sqrt { 2 } -\sqrt { 6 } -2 }{ 25-25 } \)
\(=\frac { 5\sqrt { 3 } +5\sqrt { 2 } -\sqrt { 6 } -2 }{ 23 } \)

\({ 3 }^{ \frac { 1 }{ 2 } },{ 4 }^{ \frac { 1 }{ 3 } },{ 6 }^{ \frac { 1 }{ 4 } }={ 3 }^{ \frac { 6 }{ 12 } },{ 4 }^{ \frac { 4 }{ 12 } },{ 6 }^{ \frac { 3 }{ 12 } }\)
=\({ \left( { 3 }^{ 6 } \right) }^{ \frac { 6 }{ 12 } },{ \left( { 4 }^{ 4 } \right) }^{ \frac { 4 }{ 12 } },{ \left( { 6 }^{ 3 } \right) }^{ \frac { 3 }{ 12 } }\)
= \({ 729 }^{ \frac { 1 }{ 12 } },{ 256 }^{ \frac { 1 }{ 12 } },{ 216 }^{ \frac { 1 }{ 12 } }\)
In ascending order = \({ 216 }^{ \frac { 1 }{ 12 } },{ 256 }^{ \frac { 1 }{ 12 } },{ 729 }^{ \frac { 1 }{ 12 } }\)
i.e., \(\sqrt [ 4 ]{ 6 } ,\sqrt [ 3 ]{ 4 } ,\sqrt { 3 } \)

\(\frac { 1 }{ \sqrt { 7 } -2 } \)
\(\frac { 1 }{ \sqrt { 7 } -2 } =\frac { 1 }{ \sqrt { 7 } -2 } \times \frac { \sqrt { 7 } +2 }{ \sqrt { 7 } +2 } \)
\(\\ =\frac { \sqrt { 7 } +2 }{ 7-4 } =\frac { \sqrt { 7 } +2 }{ 3 } \)

\(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } } \)=\(\frac { 2 }{ 7 } \) which is a rational number. 

Yes! We can predict the decimal expansions of \(\frac { 2 }{ 7 } ,\frac { 3 }{ 7 } ,\frac { 4 }{ 7 } ,\frac { 5 }{ 7 } ,\frac { 6 }{ 7 } \) without actually doing the long division as follows:
To predict the decimal expansion of 2/7, locate when the remainder becomes 2 and respective quotient.Then write the new quotient beginning from there using the repeating digits 1,4,2,8,5,7.
\(\frac { 1 }{ 7 } =0.\overline { 142857 } \)
Similarly,
\(\frac { 2 }{ 7 } =0.\overline { 285714 }\)
\( \\ \frac { 3 }{ 7 } =0.\overline { 428571 }\)
\( \\ \frac { 4 }{ 7 } =0.\overline { 571428 } \)
\(\\ \frac { 5 }{ 7 } =0.\overline { 714285 } \)
\(\\ \frac { 6 }{ 7 } =0.\overline { 857142 } \)

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.

\(\frac { 1 }{ x } =\frac { 1 }{ 3+2\sqrt { 2 } } =\frac { 1 }{ 3+2\sqrt { 2 } } \times \frac { 3-2\sqrt { 2 } }{ 3-2\sqrt { 2 } } \)
\(\\ =\frac { 3-2\sqrt { 2 } }{ { \left( 3 \right) }^{ 2 }-{ \left( 2\sqrt { 2 } \right) }^{ 2 } } =\frac { 3-2\sqrt { 2 } }{ 9-8 } \)
\(\\ =\frac { 3-2\sqrt { 2 } }{ 1 } =3-2\sqrt { 2 } \)
\(\\ x+\frac { 1 }{ x } =\left( 3+2\sqrt { 2 } \right) +\left( 3-2\sqrt { 2 } \right) =6\)
\(\\ { \left( x+\frac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3(x)\left( \frac { 1 }{ x } \right) \left( x+\frac { 1 }{ x } \right)\)
\( \\ { \left( 6 \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3(6)\)
\(\\ 216={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +18\)
\(\\ { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =198\)

\((3)(2)+3\sqrt { 2 } +(\sqrt { 3 } )(2)+(\sqrt { 3 } )(\sqrt { 2 } )\)
\(\\ =6+3\sqrt { 2 } +2\sqrt { 3 } +\sqrt { (3)(2) } \)
\(\\ 6+3\sqrt { 2 } +2\sqrt { 3 } +\sqrt { 6 } \)

Let x = \(1.\overline { 32 } \)
x = 1.3222...
10x = 13.222...       ...(1)
100x = 132.222...   ....(2)
Subtracting (1) from (2), we get
90x = 119
x = \(\frac { 119 }{ 90 } \)          ....(3)
Let x = \(0.\overline { 35 } \)
Then x = 0.35 35 35...(4)
100x = 35.35 35 35...(5)
Subtracting (4) from (5), we get
99x = 35
x = 35/99
\(1.\overline { 32 } +0.\overline { 35 } =\frac { 119 }{ 90 } +\frac { 35 }{ 99 } \)
\(\\ =\frac { 1309+350 }{ 990 } =\frac { 1659 }{ 990 } =\frac { 553 }{ 330 } \)
Here, p = 553, q =3 30(\(\neq 0\))

Let x = \(0.\overline { 235 } \)
x = 0.235 235 235 235...    ...(1)
1000x = 235.235 235 235 235...    ...(2)
Subtracting (1) from (2), we get
999x=235
\(x=\frac { 235 }{ 999 } \)
Here, p = 235, q = 999(\(\neq 0\))

\(\frac { 1 }{ 3 } \)=0.3333...
\(\frac { 1 }{ 2 } \)=0.5
Hence two irrational numbers between \(\frac { 1 }{ 3 } \) and \(\frac { 1 }{ 2 } \) can be taken as 0.343443444... and 0.353553555...

\(\frac { 3 }{ 7 } =\frac { 30 }{ 70 } \)
\(\\ \frac { 5 }{ 7 } =\frac { 50 }{ 70 } \)
30<31<32<33<34<35
\(\frac { 30 }{ 70 } <\frac { 31 }{ 70 } <\frac { 32 }{ 70 } <\frac { 33 }{ 70 } <\frac { 34 }{ 70 } <\frac { 50 }{ 70 } \)
So four rational numbers between \(\frac { 3 }{ 7 } \) and \(\frac { 5 }{ 7 } \)
\(\frac { 31 }{ 70 } ,\frac { 32 }{ 70 } ,\frac { 33 }{ 70 } \) and \(\frac { 34 }{ 70 } \)
\(\frac { 31 }{ 70 } ,\frac { 16 }{ 35 } ,\frac { 33 }{ 70 } \)and \(\frac { 17 }{ 35 } \)

\(\frac { 3 }{ 5 } =\frac { 3\times 8 }{ 5\times 8 } =\frac { 24 }{ 40 } \)
\(\\ \frac { 7 }{ 8 } =\frac { 7\times 5 }{ 8\times 5 } =\frac { 35 }{ 40 } \)
\(\\ \because 24<25<26<27<35\)
\(\\ \therefore \frac { 24 }{ 40 } <\frac { 25 }{ 40 } <\frac { 26 }{ 40 } <\frac { 27 }{ 40 } <\frac { 35 }{ 40 } \)
Hence, three rational numbers between  \(\frac { 3 }{ 5 } \) and \(\frac { 7 }{ 8 } \) can be taken as
\(\frac { 25 }{ 40 } ,\frac { 26 }{ 40 } \) and \(\frac { 27 }{ 40 } \)
\(\frac { 5 }{ 8 } ,\frac { 13 }{ 20 } \) and \(\frac { 27 }{ 40 } \) 

\(\frac{1}{\sqrt{2}+1}\)=\(\frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } \) = √2-1
= 1.414-1
= 0.414

Let x = \(0.\bar{6}\)
x = 0.66666...(i)
multiplying 10 on both the sides, we get
10x = 6.6666...(ii)
From (ii)-(i), we get
9x = 6.0
x = \(\frac{6}{9}=\frac{2}{3}\)

3

\(\frac { 27 }{ 225 } \)

\(2-\sqrt { 3 } \)

\(6 \sqrt{5} \times 2 \sqrt{5}=6 \times 2 \times \sqrt{5} \times \sqrt{5}=12 \times 5=60\)

610/33

Since we do not know what 0 3 . is , let us call it ‘x’ and so
x = 0.3333
Now here is where the trick comes in. Look at
10 x = 10 x (0.333...) = 3.333
Now, 3.3333 = 3 + x, since x = 0.3333
Therefore, 10 x = 3 + x
Solving for x, we get
\(9 x=3, \text { i.e., } x=\frac{1}{3}\)

7/6,4/3,3/2,5/3 and 11/6

0.125, 0.15

\(x=3-2\sqrt { 2 } \Rightarrow \frac { 1 }{ x } =3+2\sqrt { 2 } \)
\({ \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) }^{ 2 }=8\)
\(\Rightarrow \sqrt { x } +\frac { 1 }{ \sqrt { x } } =\pm 2\sqrt { 2 } \)

\(={ \left( { x }^{ { a }^{ 2 }-{ b }^{ 2 } } \right) }^{ \frac { 1 }{ a+b } }.{ \left( { x }^{ { b }^{ 2 }-{ c }^{ 2 } } \right) }^{ \frac { 1 }{ b+c } }.{ \left( { x }^{ { c }^{ 2 }-{ a }^{ 2 } } \right) }^{ \frac { 1 }{ c+a } }\)
\(={ x }^{ \frac { { a }^{ 2 }-{ b }^{ 2 } }{ a+b } }.{ x }^{ \frac { { b }^{ 2 }-{ c }^{ 2 } }{ a+b } }.{ x }^{ \frac { { c }^{ 2 }-{ a }^{ 2 } }{ c+a } }\)
\(={ x }^{ a-b }.{ x }^{ b-c }.{ x }^{ c-a }\)
\(={ x }^{ 0 }\)

\(\frac { 4 }{ { \left( 216 \right) }^{ -\frac { 2 }{ 3 } } } +\frac { 1 }{ { \left( 216 \right) }^{ -\frac { 3 }{ 4 } } } +\frac { 2 }{ { \left( 216 \right) }^{ -\frac { 1 }{ 5 } } } \)
\(=\frac { 4 }{ { \left( { 6 }^{ 3 } \right) }^{ -\frac { 2 }{ 3 } } } +\frac { 1 }{ { \left( { 4 }^{ 4 } \right) }^{ -\frac { 3 }{ 4 } } } +\frac { 2 }{ { \left( { 3 }^{ 5 } \right) }^{ -\frac { 1 }{ 5 } } } \)
\(=\frac { 4 }{ { 6 }^{ -2 } } +\frac { 1 }{ { 4 }^{ -3 } } +\frac { 2 }{ { 3 }^{ -1 } } \)
\(=\frac { 4 }{ { 36 }^{ -1 } } +\frac { 4 }{ { 64 }^{ -1 } } +\frac { 2 }{ { 3 }^{ -1 } } \)
= 4 x 36 + 1 x 64 + 2 x 3
= 214

\(\frac { { 3 }^{ 30 }+{ 3 }^{ 29 }+{ 3 }^{ 28 } }{ { 3 }^{ 31 }+{ 3 }^{ 30 }-{ 3 }^{ 29 } } =\frac { { 3 }^{ 38 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }+1 \right) }{ { 3 }^{ 29 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }-1 \right) } \)
\(=\frac { \left( 9+3+1 \right) }{ 3\left( 9+3-1 \right) } \)
\(=\frac { 13 }{ 3\times 11 } =\frac { 13 }{ 33 } \)