Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

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

(0.3)^{[{200 - 146}/(3 × 3 × 3) - 3]}
= (0.3)^{[{54}/(27) - 3]}
= (0.3)^{[2 - 3]}
= (0.3)^{[- 1]}
= $$\frac{1}{{0.3}}$$
= $$\frac{{10}}{3}$$

$$\eqalign{ & \frac{{2\sqrt {10} }}{{\sqrt 5 + \sqrt 2 - \sqrt 7 }} - \sqrt {\frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}}} - \frac{3}{{\sqrt 7 - 2}} \cr & = \frac{{2\sqrt {10} \left( {\sqrt 5 + \sqrt 2 + \sqrt 7 } \right)}}{{{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2} - {{\left( {\sqrt 7 } \right)}^2}}} - \sqrt {\frac{{{{\left( {\sqrt 5 - 2} \right)}^2}}}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}}}} - \frac{{3\left( {\sqrt 7 + 2} \right)}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( 2 \right)}^2}}} \cr & = \frac{{2\sqrt {10} \left( {\sqrt 5 + \sqrt 2 + \sqrt 7 } \right)}}{{5 + 2 + 2\sqrt {10} - 7}} - \frac{{\left( {\sqrt 5 - 2} \right)}}{1} - \frac{{3\left( {\sqrt 7 + 2} \right)}}{3} \cr & = \sqrt 5 + \sqrt 2 + \sqrt 7 - \sqrt 5 + 2 - \sqrt 7 - 2 \cr & = \sqrt 2 \cr} $$