Solution:

 \(\displaystyle \sqrt{ 3 }(\sqrt{ 7 } - \sqrt{ 3 })\)

= \(\displaystyle \sqrt{ 3 \times 7 } - \sqrt{ 3 \times 3 }\)

= \(\displaystyle \sqrt{ 3 \times 7 } - \sqrt{ 3 \times 3 }\)

= \(\displaystyle \sqrt{ 21 } - \sqrt{ 9 }\)

= \(\displaystyle \sqrt{ 21 } - 3\)

= \(\displaystyle - 3 + \sqrt{ 21 } \)

[When we write a rational number and an irrational number together, it is customary to write the rational number first.]

Solution:

 \(\displaystyle (\sqrt{ 5 } - \sqrt{ 7 })\sqrt{ 2 })\)

= \(\displaystyle \sqrt{ 5 \times 2 } - \sqrt{ 7 \times 2 }\)

= \(\displaystyle \sqrt{ 10 } - \sqrt{ 14 }\)

Solution:

 \(\displaystyle 3 \sqrt{ 2 }(4 \sqrt{ 3 } - \sqrt{ 2 }) - \sqrt{ 3 })(4 \sqrt{ 3 } - \sqrt{ 2 })\)

= \(\displaystyle 3 \times 4 \times \sqrt{ 2 } \times \sqrt{ 3 } - 3 \times \sqrt{ 2 } \times \sqrt{ 2 } - 4 \times \sqrt{ 3 } \times \sqrt{ 3 } + \sqrt{ 3 } \times \sqrt{ 2 }\)

= \(\displaystyle 13\sqrt{ 6 } - 3 \times 2 - 4 \times 3\)

= \(\displaystyle 13\sqrt{ 6 } - 6 - 12\)

= \(\displaystyle 13\sqrt{ 6 } - 18\)

= \(\displaystyle - 18 + 13\sqrt{ 6 }\)

Solution:

 \(\displaystyle \frac { 1 }{ \sqrt { 7 } + \sqrt { 2 }}\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 7 } - \sqrt { 2 }\),

 \(\displaystyle \frac { 1 }{ \sqrt { 7 } + \sqrt { 2 }} \times \displaystyle \frac {\sqrt { 7 } - \sqrt { 2 }} {\sqrt { 7 } - \sqrt { 2 }}\)

= \(\displaystyle \frac { 1(\sqrt { 7 } - \sqrt { 2 })}{(\sqrt { 7 } + \sqrt { 2 }){(\sqrt { 7 } + \sqrt { 2 })} }\)

= \(\displaystyle \frac { \sqrt { 7 } - \sqrt { 2 }}{(\sqrt { 7 })^2 - {(\sqrt { 5 })^2} }\) ... [∵ (a − b)(a + b) = a² − b² ]

= \(\displaystyle \frac { \sqrt { 7 } - \sqrt { 2 }}{7 - 2}\)

= \(\displaystyle \frac { \sqrt { 7 } - \sqrt { 2 }}{5}\)

Solution:

 \(\displaystyle \frac { 3 }{ 2 \sqrt { 5 } - 3 \sqrt { 2 }}\)

Multiplying the numerator and denominator by \(\displaystyle 2 \sqrt { 5 } + 3 \sqrt { 2 }\),

 \(\displaystyle \frac { 3 }{ 2 \sqrt { 5 } - 3 \sqrt { 2 }} \times \frac {2 \sqrt { 5 } + 3 \sqrt { 2 }}{2 \sqrt { 5 } + 3 \sqrt { 2 }}\)

= \(\displaystyle \frac { 3(2 \sqrt { 5 } + 3 \sqrt { 2 })}{(2 \sqrt { 5 } - 3 \sqrt { 2 }) {(2 \sqrt { 5 } + 3 \sqrt { 2 })} }\)

= \(\displaystyle \frac { 3(2 \sqrt { 5 } + 3 \sqrt { 2 })}{(2 \sqrt { 5 })^2 - {(3 \sqrt { 2 })^2} }\) ... [∵ (a − b)(a + b) = a² − b² ]

= \(\displaystyle \frac { 3(2 \sqrt { 5 } + 3 \sqrt { 2 })}{4 \times 5 - 9 \times 2}\)

= \(\displaystyle \frac { 3(2 \sqrt { 5 } + 3 \sqrt { 2 })}{20 - 18}\)

= \(\displaystyle \frac { 3(2 \sqrt { 5 } + 3 \sqrt { 2 })}{2}\)

Solution:

 \(\displaystyle \frac { 4 }{ 7 + 4 \sqrt { 3 }}\)

Multiplying the numerator and denominator by \(\displaystyle 7 - 4 \sqrt { 3 }\),

 \(\displaystyle \frac { 4 }{ 7 + 4 \sqrt { 3 }} \times \displaystyle \frac {7 - 4 \sqrt { 3 }} {7 - 4 \sqrt { 3 }}\)

= \(\displaystyle \frac { 4(7 - 4 \sqrt { 3 })}{(7 + 4 \sqrt { 3 })(7 - 4 \sqrt { 3 }) }\)

= \(\displaystyle \frac { 4(7 - 4 \sqrt { 3 })}{7^2 - {(4 \sqrt { 3 })^2} }\) ... [∵ (a + b)(a − b) = a² − b² ]

= \(\displaystyle \frac { 28 - 16 \sqrt { 3 }}{49 - 16 \times 3}\)

= \(\displaystyle \frac { 28 - 16 \sqrt { 3 }}{49 - 48}\)

= \(\displaystyle \frac { 28 - 16 \sqrt { 3 }}{1}\)

= \(\displaystyle 28 - 16 \sqrt { 3 }\)

Solution:

 \(\displaystyle \frac { \sqrt { 5 } - \sqrt { 3 } }{ \sqrt { 5 } + \sqrt { 3 }}\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 5 } - \sqrt { 3 }\),

 \(\displaystyle \frac { \sqrt { 5 } - \sqrt { 3 } }{ \sqrt { 5 } + \sqrt { 3 }} \times \frac {\sqrt { 5 } - \sqrt { 3 }} {\sqrt { 5 } - \sqrt { 3 }}\)

= \(\displaystyle \frac { (\sqrt { 5 } - \sqrt { 3 })(\sqrt { 5 } - \sqrt { 3 })}{(\sqrt { 5 } + \sqrt { 3 })(\sqrt { 5 } - \sqrt { 3 }) }\)

= \(\displaystyle \frac { (\sqrt { 5 } - \sqrt { 3 })^2}{(\sqrt { 5 })^2 - {(\sqrt { 3 })^2} }\) ... [∵ (a + b)(a − b) = a² − b² ]

= \(\displaystyle \frac { (\sqrt { 5 })^2 - 2 \times \sqrt { 5 } \times \sqrt { 3 } + (\sqrt { 3 })^2}{5 - 3 }\) ... [∵ (a − b)² = a² − 2ab + b²]

= \(\displaystyle \frac { 8 - 2 \sqrt { 15 }}{2}\)

= \(\displaystyle \frac { 2(4 - \sqrt { 15 })}{2}\)

= \(\displaystyle 4 - \sqrt { 15 }\)