Solution:
$$\eqalign{ \left( {4 + \sqrt 7 } \right) \cr = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$

Solution:
$$\eqalign{ {\left( {\sqrt {11} + \sqrt 5 } \right)^2} = 16 + 2\sqrt {55} \cr {\left( {\sqrt {14} + \sqrt 2 } \right)^2} = 16 + 2\sqrt {28} \cr {\left( {\sqrt 8 + \sqrt 8 } \right)^2} = 16 + 2\sqrt {64} \cr {\text{It is clear that }}16 + 2\sqrt {64} {\text{ is largest}} \cr {\text{So, }}\left( {\sqrt 8 + \sqrt 8 } \right){\text{ will be largest}}{\text{.}} \cr} $$

Solution:
$$\eqalign{ {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr = {10^{150 - 146}} \cr = {10^4} \cr = 10000 \cr} $$

Solution:
$$\eqalign{ x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr \Rightarrow x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr \Rightarrow x = \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{2} \cr {\text{Similarly}} \cr y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr \Rightarrow y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{2} \cr {\text{Now, }}x + y \cr = \frac{{5 + 3 + 2\sqrt {15} + 5 + 3 - 2\sqrt {15} }}{2} \cr = \frac{{16}}{2} \cr = 8 \cr} $$

Solution:
$$\eqalign{ {\text{Given}} \cr x = \sqrt {a\root 3 \of {b\sqrt {a\root 3 \of b } } } \,.....\,\infty \,.\,.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr {\text{On squaring both sides}} \cr \Rightarrow {x^2} = a\root 3 \of {b\sqrt {a\root 3 \of b } } \,.....\,\infty \cr {\text{On cubing both sides}} \cr \Rightarrow {x^6} = {a^3}b\,\sqrt {a\root 3 \of {b\sqrt {a\root 3 \of b } } } \,.....\,\infty \cr \Rightarrow {x^6} = {a^3}b\,x{\text{ from equation }}\left( {\text{i}} \right) \cr {\text{On dividing above equation by }}x{\text{ we get}} \cr \Rightarrow \frac{{{x^6}}}{x} = \frac{{{a^3}bx}}{x} \cr \Rightarrow {x^5} = {a^3}b \cr \Rightarrow x = \root 5 \of {{a^3}b} \cr} $$

Solution:
$$\eqalign{ \sqrt {8 - 2\sqrt {15} } \cr = \sqrt {5 + 3 - 2 \times \sqrt 5 \times \sqrt 3 } \cr = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2 \times \sqrt 5 \times \sqrt 3 } \cr = \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \cr = \left( {\sqrt 5 - \sqrt 3 } \right) \cr} $$

Solution:
$$\eqalign{ \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr = \left[ {8 - {{\left( {\frac{{{2^{2 \times \frac{9}{4}}}\sqrt {{2^{1 + 2}}} }}{{2\sqrt {\frac{1}{4}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr = \left[ {8 - {{\left( {\frac{{{2^{\frac{9}{2}}}{{.2}^{\frac{3}{2}}}}}{{2 \times \frac{1}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr = \left[ {8 - {{\left( {{2^{\frac{{12}}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr = \left[ {8 - \left( {{2^{6 \times \frac{1}{2}}}} \right)} \right] \cr = \left[ {8 - \left( {{2^3}} \right)} \right] \cr = \left[ {8 - 8} \right] \cr = 0 \cr} $$