\(x=1+\sqrt { 2 } \)
\(\Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 1+\sqrt { 2 } } \times \frac { 1-\sqrt { 2 } }{ 1-\sqrt { 2 } } =\frac { 1-\sqrt { 2 } }{ 1-2 } \)
\(=\sqrt { 2 } -1\)
\(x-\frac { 1 }{ x } =\left( 1+\sqrt { 2 } \right) -\left( \sqrt { 2 } -1 \right) \)
\(=1+\sqrt { 2 } -\sqrt { 2 } +1=2\)
\({ \left( x-\frac { 1 }{ x } \right) }^{ 3 }\)= 2³ = 8

\(\frac{5}{7}\) =\(0.\overline{714285}\)
\(\frac{9}{11}=0.\overline{81}\)
Hence three irrational numbers between \(\frac{5}{7}\) and \(\frac{9}{11}\) can be:
0.727227222...
0.737337333...
0.747447444...

\(\frac { 25 }{ 99 } \)

\(\frac { 3 }{ 13 } \)= 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 \(\frac { 27 }{ 7 } \)

\(x=\frac { 1 }{ 2-\sqrt { 3 } } \)
\(\Rightarrow x=\frac { 1 }{ 2-\sqrt { 3 } } \times \frac { \left( 2+\sqrt { 3 } \right) }{ \left( 2+\sqrt { 3 } \right) } \)
\(\Rightarrow \quad \frac { 2+\sqrt { 3 } }{ 4-3 } =2+\sqrt { 3 } \)
⇒ (x-2) = √3
⇒ (x-2)² = (√3)² = 3
x² - 4x + 4 = 3
x²- 4x + 4 - 3 = 0
x² - 4x + 1 = 0
x³ - 2x² - 7x + 5
x(x² - 4x + 1)+ 2(x² - 4x + 1) + 3= X0 + 2X0 + 3 = 3

\(\frac { { \left( { 3 }^{ 2 } \right) }^{ n+1 }\times { 3 }^{ n }-{ \left( { 3 }^{ 3 } \right) }^{ n } }{ { 3 }^{ 3m }\times { 2 }^{ 3 } } =\frac { 1 }{ { 3 }^{ 6 } } \)
⇒ \(\frac { { 3 }^{ 2n+2 }\times { 3 }^{ n }-{ 3 }^{ n } }{ { 3 }^{ 3m }\times 8 } =\frac { 1 }{ { 3 }^{ 6 } } \)
ஃ 3n - 3m = -6
ஃ m - n = -2

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

Let x = \(2.\overline { 218 } \)
x = 2.2181818...
10x = 22.181818...   ...(1)
1000x = 2218.181818...  ...(2)
Subtracting (1) from (2), we get
990x = 2196
\(x=\frac { 2196 }{ 990 } =\frac { 122 }{ 55 } \)
Here, p = 122, q = 55(\(\neq 0\))

\(\frac { 1 }{ 5 } =\frac { 1\times 6 }{ 5\times 6 } =\frac { 6 }{ 30 } =\frac { 6\times 10 }{ 30\times 10 } =\frac { 60 }{ 300 } \) 
\(\\ \frac { 1 }{ 6 } =\frac { 1\times 5 }{ 6\times 5 } =\frac { 5 }{ 30 } =\frac { 5\times 30 }{ 30\times 10 } =\frac { 50 }{ 300 }\)
\( \\ \because 50<51<52<53<54<60\)
\(\\ \therefore \frac { 50 }{ 300 } <\frac { 51 }{ 300 } <\frac { 52 }{ 300 } <\frac { 53 }{ 300 } <\frac { 54 }{ 300 } <\frac { 60 }{ 300 } \)
Hence, four rational numbers between \(\frac { 1 }{ 5 } \) and \(\frac { 1 }{ 6 } \) can be taken as
\(\frac { 51 }{ 300 } ,\frac { 52 }{ 300 } ,\frac { 53 }{ 300 } \)and \(\frac { 54 }{ 300 } \)
or \(\frac { 17 }{ 100 } ,\frac { 13 }{ 75 } ,\frac { 53 }{ 300 } \)and \(\frac { 9 }{ 50 } \)

\(\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } \)
\(\frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 1.414 }{ 2 } \)= 0.707
ஃ \(\frac { 1 }{ \sqrt { 2 } } +\pi =0.707+3.141\)
= 3.848

5√2(3 + √2)(5 + √2)
= 5√2[17 + 8√2] = 85√2 + 80

8√3 - 2√3 + 4√3 = √3(8 - 2 + 4)
= 10√3

LCM of 5 and 7 is 35
\(\frac{3}{5}=\frac{3}{5}\times\frac{7}{7}=\frac{21}{35}\)
and \(\frac{5}{7}=\frac{5}{7}\times\frac{5}{5}=\frac{25}{35}\)
so, \(\frac{21}{35}<\frac{22}{35}<\frac{23}{35}<\frac{24}{35}<\frac{25}{35}\)
The required three rational numbers are \(\frac{22}{35},\frac{23}{35}\) and \(\frac{24}{35}\)

2

80

\(a=\frac { 138 }{ 59 } ,b=0\)

\((i)\ 10+5\sqrt { 5 } +2\sqrt { 7 } +\sqrt { 35 } \)
\((ii)\ 20\)
\((iii)\ 10+2\sqrt { 21 } \)
\(\\ (iv)\ 4\)

0.414

\(\frac{1}{7}=0.142857142857 \ldots=0 . \overline{142857}\) and \(\frac{2}{7}=0.28571428571428 \ldots=0 . \overline{285714}\)
Here, the two decimal expansions are non-terminating recurring.
Hence, 1/7​ and 2/7​ are two rational numbers.
We know, between any two rational numbers, there are infinitely many irrational numbers.
An irrational number has non-terminating non-recurring decimal expansions.
Then an irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\) is 0.15015001500015.
Similarly, 0.21020020002... is another irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\)

\(\frac { 1 }{ 1+{ x }^{ a-b } } +\frac { 1 }{ 1+{ x }^{ b-a } } =\frac { { x }^{ b } }{ { x }^{ b }+{ x }^{ a } } +\frac { { x }^{ a } }{ { x }^{ a }+{ x }^{ b } } \)
\(=\frac { { x }^{ b }+{ x }^{ a } }{ { x }^{ a }+{ x }^{ b } } \)

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