Answer: Option C. -> $$\sqrt 5 - \sqrt 6 $$
$$\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}$$
$$ \Rightarrow \frac{{\left( {1 + \sqrt 2 } \right)\left( {\sqrt 5 - \sqrt 3 } \right) + \left( {1 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)}}$$
$$ \Rightarrow \frac{{\sqrt 5 - \sqrt 3 + \sqrt {10} - \sqrt 6 + \sqrt 5 + \sqrt 3 - \sqrt {10} - \sqrt 6 }}{{5 - 3}}$$
$$\eqalign{
& \Rightarrow \frac{{2\sqrt 5 - 2\sqrt 6 }}{2} \cr
& \Rightarrow \frac{{2\left( {\sqrt 5 - \sqrt 6 } \right)}}{2} \cr
& \Rightarrow \sqrt 5 - \sqrt 6 \cr} $$

Answer: Option C. -> $$\left\{ {\frac{1}{2}{{\left( {\sqrt 7 + 1} \right)}^2}} \right\}$$
$$\eqalign{
& \left( {4 + \sqrt 7 } \right) \cr
& = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr
& = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr
& = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr
& = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$

Answer: Option D. -> $$\frac{{17}}{{15}}$$
$$\eqalign{
& {\text{According to question,}} \cr
& \Rightarrow {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr
& \Rightarrow {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr
& \Rightarrow x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr
& x + y = 4......{\text{(i)}} \cr
& {\text{ }}x - y = \frac{1}{4}.....(ii) \cr
& {\text{Solve the equation of (i) and (ii)}} \cr
& x = \frac{{17}}{8}, \cr
& y = \frac{{15}}{8}, \cr
& \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$

Answer: Option C. -> 176
$$\eqalign{
& \left( {\sqrt 3 + 1} \right)\left( {10 + \sqrt {12} } \right)\left( {\sqrt {12} - 2} \right)\left( {5 - \sqrt 3 } \right) \cr
& \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {10 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 2} \right)\left( {5 - \sqrt 3 } \right) \cr} $$
$$ \Rightarrow \left( {\sqrt 3 + 1} \right) \times $$   $$2\left( {5 + \sqrt 3 } \right) \times $$   $$2\left( {\sqrt 3 - 1} \right)$$  $$\left( {5 - \sqrt 3 } \right)$$
$$\eqalign{
& \Rightarrow 4\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)\left( {5 + \sqrt 3 } \right)\left( {5 - \sqrt 3 } \right) \cr
& \Rightarrow 4\left[ {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} \right]\left[ {{{\left( 5 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right] \cr
& \Rightarrow 4 \times 2 \times 22 \cr
& \Rightarrow 176 \cr} $$

Answer: Option B. -> 9.898
$$\eqalign{
& {\text{2}}\sqrt {50} {\text{ + }}\sqrt {18} - \sqrt {72} \cr
& \Rightarrow {\text{2}} \times {\text{5}}\sqrt 2 {\text{ + 3}}\sqrt 2 - 6\sqrt 2 \cr
& \Rightarrow 13\sqrt 2 - 6\sqrt 2 \cr
& \Rightarrow 7\sqrt 2 \cr
& \Rightarrow 7 \times 1.414 \cr
& \Rightarrow 9.898 \cr} $$

Answer: Option B. -> 0
$$\eqalign{
& {\text{Let }}a = 55, \cr
& \,\,\,\,\,\,\,\,\,\,\,b = 17, \cr
& \,\,\,\,\,\,\,\,\,\,\,c = - 72 \cr
& a + b + c \cr
& = 55 + 17 - 72 \cr
& = 0 \cr
& \therefore {a^3} + {b^3} + {c^3} - 3abc = 0 \cr
& \left( {a + b + c} \right) = 0 \cr
& {\text{Answer is }}0. \cr} $$

Answer: Option B. -> 1
$$\eqalign{
& {\text{Given expression, }} \cr
& \frac{1}{{1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}}} + \frac{1}{{1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}}} + \frac{1}{{1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}}} \cr} $$
  $$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$    $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^{b}} + {x^c}} \right)}}$$
$$\eqalign{
& = \frac{{\left( {{x^a} + {x^b} + {x^c}} \right)}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} \cr
& = 1 \cr} $$

Answer: Option C. -> 1
$$\eqalign{
& \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr
& = \frac{1}{{1 + \frac{{{a^n}}}{{{a^m}}}}} + \frac{1}{{1 + \frac{{{a^m}}}{{{a^n}}}}} \cr
& = \frac{{{a^m}}}{{{a^m} + {a^n}}} + \frac{{{a^n}}}{{{a^m} + {a^n}}} \cr
& = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr
& = 1 \cr} $$

Answer: Option A. -> 1
$$\eqalign{
& {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^0} \cr
& = 1 \cr} $$

Answer: Option D. -> 7
$$\eqalign{
& {\text{ }}{{\text{2}}^{n - 1}}{\text{ + }}{{\text{2}}^{n + 1}}{\text{ = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}\left( {1 + {2^2}} \right){\text{ = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}} \times {\text{5 = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}{\text{ = }}\frac{{320}}{5}{\text{ = 64}} \cr
& \Rightarrow {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^6} \cr
& \Rightarrow n - 1 = 6 \cr
& \Rightarrow n = 7 \cr} $$