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

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

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

$$\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {3x + 2y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \Rightarrow {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & \Rightarrow {\left( 5 \right)^2} \cr & \Rightarrow 25 \cr} $$

$$\eqalign{ & {\text{The exponential form of }} \cr & \sqrt {\sqrt 2 \times \sqrt 3 } \cr & = \sqrt {{6^{\frac{1}{2}}}} \cr & = {\left( {{6^{\frac{1}{2}}}} \right)^{\frac{1}{2}}} \cr & = {6^{\frac{1}{4}}} \cr} $$

$$\eqalign{ & {\text{Expression,}} \cr & {\text{ = }}\frac{{{{\left( {10} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( {2 \times 5} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( 2 \right)}^{100}} \times {{\left( 5 \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = {2^{100}} \times \frac{{{5^{100}}}}{{{5^{75}}}} \cr & = {2^{100}} \times {5^{\left( {100 - 75} \right)}}.....\left[ {\because \frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right] \cr & = {2^{100}} \times {5^{25}} \cr & = {2^{25}} \times {5^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & = {\left( {10} \right)^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {b^m} = a{b^m}} \right] \cr} $$

LCM of 2, 3, 4, 6 is 12
$$\sqrt 2 = $$  $${2^{\frac{1}{2}}} = $$  $${2^{\left( {\frac{1}{2} \times \frac{6}{6}} \right)}} = $$   $${2^{\frac{6}{{12}}}} = $$  $${\left( {{2^6}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {64} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {64} $$
$$\root 6 \of 3 = $$  $${3^{\frac{1}{6}}} = $$  $${3^{\left( {\frac{1}{6} \times \frac{2}{2}} \right)}} = $$   $${3^{\frac{2}{{12}}}} = $$  $${\left( {{3^2}} \right)^{\frac{1}{{12}}}} = $$  $${\left( 9 \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of 9 $$
$$\root 3 \of 4 = $$  $${4^{\frac{1}{3}}} = $$  $${4^{\left( {\frac{1}{3} \times \frac{4}{4}} \right)}} = $$  $${4^{\frac{4}{{12}}}} = $$  $${\left( {{4^4}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {256} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {256} $$
$$\root 4 \of 5 = $$  $${5^{\frac{1}{4}}} = $$  $${5^{\left( {\frac{1}{4} \times \frac{3}{3}} \right)}} = $$   $${5^{\frac{3}{{12}}}} = $$  $${\left( {{5^3}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {125} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {125} $$
Clearly $$\root {12} \of {256} \,\,i.e.,\root 3 \of 4 $$     is the greatest.

$$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr} $$

$$\eqalign{ & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {154 + \sqrt {225} } } } } }}{{\root 3 \of 8 }} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {169} } } } }}{2} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {25 + \sqrt {121} } } }}{2} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {36} } }}{2} \cr & \Rightarrow \frac{{\sqrt {16} }}{2} \cr & \Rightarrow \frac{4}{2} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\left( {32 \times {{10}^{ - 5}}} \right)^{ 2}} \times 64 \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) = {10^?} \cr & \Rightarrow {\left( {{2^5} \times {{10}^{ - 5}}} \right)^{ 2}} \times {2^6} \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) \cr & \,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right] \cr & \Rightarrow \frac{{{2^{10}} \times {{10}^{ - 10}} \times {2^6}}}{{{2^{16}} \times {{10}^{ - 4}}}} \cr & \,\,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow \frac{{{2^{16}} \times {{10}^4}}}{{{2^{16}} \times {{10}^{10}}}} \cr & \,\,\,\,\,\,\,\, = {10^?}.....\left[ {{a^{ - m}} = \frac{1}{{{a^m}}}} \right] \cr & \Rightarrow {10^{4 - 10}} = {10^?} \cr & \Rightarrow {10^{ - 6}} = {10^?} \cr & \Rightarrow ? = - 6 \cr} $$

$$\eqalign{ & 8 = {7^d} \cr & \,\,\,\,\,\, = {\left( {{6^c}} \right)^d} \cr & \,\,\,\,\,\, = {\left( {{5^b}} \right)^{cd}} \cr & \,\,\,\,\,\, = {5^{bcd}} \cr & \,\,\,\,\,\, = {\left( {{4^a}} \right)^{bcd}} \cr & \,\,\,\,\,\, = {4^{abcd}} \cr & \Rightarrow {4^{abcd}} = 8 \cr & \Rightarrow {\left( {{2^2}} \right)^{abcd}} = {2^3} \cr & \Rightarrow 2abcd = 3 \cr & \Rightarrow abcd = \frac{3}{2} \cr} $$

$$\eqalign{ & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{2 + \sqrt 3 - \sqrt 3 - 1}} \cr & \Rightarrow {6^2} + {7^2} + {8^2} + {9^2} + {10^2} \cr & \Rightarrow 36 + 49 + 64 + 81 + 100 \cr & \Rightarrow 330 \cr} $$

$$\eqalign{ & {{\text{2}}^x}{\text{ = }}{{\text{8}}^{y + 1}} \cr & \Leftrightarrow {{\text{2}}^x}{\text{ = }}{\left( {{2^3}} \right)^{y + 1}} = {2^{\left( {3y + 3} \right)}} \cr & \Leftrightarrow x = 3y + 3 \cr & \Leftrightarrow x - 3y = 3.......(i) \cr & {9^y} = {3^{x - 9}} \cr & \Leftrightarrow {\left( {{3^2}} \right)^y}{\text{ = }}{{\text{3}}^{x - 9}} \cr & \Leftrightarrow 2y = x - 9 \cr & \Leftrightarrow x - 2y = 9......({\text{ii}}) \cr & {\text{Subtracting (i) from (ii),}} \cr & {\text{we}}\,{\text{get}}\,y = 6 \cr & {\text{Putting }}y\,{\text{ = 6 in (i),}} \cr & {\text{we get }}x{\text{ = 21}} \cr & \therefore x + y = 21 + 6 = 27 \cr} $$

$$\eqalign{ & 2\root 3 \of {32} - 3\root 3 \of 4 + \root 3 \of {500} \cr & = 2\root 3 \of {{2^3} \times 4} - 3\root 3 \of 4 + \root 3 \of {{5^3} \times 4} \cr & = 2 \times 2\root 3 \of 4 - 3\root 3 \of 4 + 5\root 3 \of 4 \cr & = 9\root 3 \of 4 - 3\root 3 \of 4 \cr & = 6\root 3 \of 4 \cr} $$

$$2\sqrt 3 = {\left( {4 \times 3} \right)^{\frac{1}{2}}} \to {12^{\frac{1}{2}}} \to {12^{\frac{2}{4}}} \to \root 4 \of {144} $$
$$2\root 4 \of 5 = \root 4 \of {\left( {5 \times 16} \right)} \to {80^{\frac{1}{4}}} \to \root 4 \of {80} $$
$$\sqrt 8 = {8^{\frac{1}{2}}} \to {8^{\frac{1}{2}}} \to \boxed{\root 4 \of {64} }\,{\text{smallest}}$$
$$3\sqrt 2 = \sqrt {18} \to {18^{\frac{1}{2}}} \to {18^{\frac{2}{4}}} \to \root 4 \of {324} $$
$$\sqrt 8 \,{\text{ is answer}}$$

$$\eqalign{ & \sqrt 2 \to {2^{\frac{1}{2}}} \to {2^{\frac{{15}}{{30}}}} = \root {30} \of {{2^{15}}} = \root {30} \of {32768} \cr & \root 3 \of 3 \to {3^{\frac{1}{3}}} \to {3^{\frac{{10}}{{30}}}} = \root {30} \of {{3^{10}}} = \root {30} \of {59049} \cr & \root 6 \of 6 \to {6^{\frac{1}{6}}} \to {6^{\frac{5}{{30}}}} = \root {30} \of {{6^5}} = \root {30} \of {7776} \cr & \root 5 \of 5 \to {5^{\frac{1}{5}}} \to {5^{\frac{6}{{30}}}} = \root {30} \of {{5^6}} = \root {30} \of {15625} \cr & {\text{So }}\root 3 \of 3 {\text{ is the greatest}}{\text{.}} \cr} $$

$$\eqalign{ & \frac{1}{{1 + \sqrt 2 + \sqrt 3 }} + \frac{1}{{1 - \sqrt 2 + \sqrt 3 }} \cr & = \frac{1}{{1 + \sqrt 3 + \sqrt 2 }} + \frac{1}{{1 + \sqrt 3 - \sqrt 2 }} \cr & = \frac{{1 + \sqrt 3 - \sqrt 2 + 1 + \sqrt 3 + \sqrt 2 }}{{{{\left( {1 + \sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{2 + 2\sqrt 3 }}{{4 + 2\sqrt 3 - 2}} \cr & = \frac{{2 + 2\sqrt 3 }}{{2 + 2\sqrt 3 }} \cr & = 1 \cr} $$

$$\eqalign{ & {21^?} \times {21^{6.5}} = {21^{12.4}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow {21^{? + 6.5}} = {21^{12.4}} \cr & \Rightarrow ? + 6.5 = 12.4 \cr & \Rightarrow ? = 12.4 - 6.5 \cr & \Rightarrow ? = 5.9 \cr} $$

$$\eqalign{ & {{\text{2}}^{0.7x}}{\text{.}}{{\text{3}}^{ - 1.25y}}{\text{ = }}\frac{{8\sqrt 6 }}{{27}} \cr & \Leftrightarrow \frac{{{{\text{2}}^{0.7x}}}}{{{{\text{3}}^{ 1.25y}}}}{\text{ = }}\frac{{{2^3}{{.2}^{\frac{1}{2}}}{{.3}^{\frac{1}{2}}}}}{{{3^3}}} \cr & \Leftrightarrow \frac{{{2^{\left( {3 + \frac{1}{2}} \right)}}}}{{{3^{\left( {3 - \frac{1}{2}} \right)}}}} = \frac{{{2^{\frac{7}{2}}}}}{{{2^{\frac{5}{2}}}}} = \frac{{{2^{3.5}}}}{{{3^{2.5}}}} \cr & \therefore 0.7x = 3.5 \Rightarrow x = \frac{{3.5}}{{0.7}}{\text{ = 5}} \cr & {\text{and }}1.25y = 2.5 \cr & \Rightarrow y = \frac{{2.5}}{{1.25}} = 2 \cr} $$

$$\eqalign{ & {\text{5}}\sqrt 5 \times {{\text{5}}^3} \div {{\text{5}}^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \Rightarrow {{\text{5}}^1} \times {5^{\frac{1}{2}}} \times {{\text{5}}^3} \div {{\text{5}}^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \Rightarrow {5^{1 + \frac{1}{2} + 3 - \left( { - \frac{3}{2}} \right)}} = {5^{a + 2}} \cr & \Rightarrow {5^{1 + \frac{1}{2} + 3 + \frac{3}{2}}} = {5^{a + 2}} \cr & \Rightarrow {5^6} = {5^{a + 2}} \cr & \Rightarrow a + 2 = 6 \cr & \Rightarrow a = 4 \cr} $$

$${x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$
  $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$

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

$$\eqalign{ & {\left( {2.89} \right)^{0.5}} = {\left( {2.89} \right)^{\frac{5}{{10}}}} \to \sqrt {2.89} \to 1.7 \cr & {\text{2}} - {\left( {0.5} \right)^2} = 2 - 0.25 \to 1.75 \cr & 1 + \frac{{0.5}}{{1 - \frac{1}{2}}} = 1 + \frac{{0.5}}{{0.5}} \to 1 + 1 \to 2 \leftarrow {\text{Greatest}} \cr & \sqrt 3 = 1.732 \cr} $$

$$\frac{1}{{\sqrt 7 - \sqrt 6 }} - $$  $$\frac{1}{{\sqrt 6 - \sqrt 5 }} + $$  $$\frac{1}{{\sqrt 5 - 2 }} - $$  $$\frac{1}{{\sqrt 8 - \sqrt 7 }} + $$  $$\frac{1}{{3 - \sqrt 8 }} = ?$$
⇒ Rationalising,
$$ \Rightarrow \frac{{\sqrt 7 + \sqrt 6 }}{{\left( {\sqrt 7 + \sqrt 6 } \right)\left( {\sqrt 7 - \sqrt 6 } \right)}} - $$     $$\frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}} \times $$   $$\frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 + \sqrt 5 } \right)}} + $$   $$\frac{{\sqrt 5 + \sqrt 4 }}{{\left( {\sqrt 5 - \sqrt 4 } \right)\left( {\sqrt 5 + \sqrt 4 } \right)}} - $$     $$\frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 - \sqrt 7 } \right)\left( {\sqrt 8 + \sqrt 7 } \right)}} + $$     $$\frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 + \sqrt 8 } \right)\left( {\sqrt 9 - \sqrt 8 } \right)}}$$
$$ \Rightarrow \frac{{\sqrt 7 + \sqrt 6 }}{1} - $$   $$\frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{1} + $$   $$\frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{1} - $$   $$\frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{1} + $$   $$\frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{1}$$
$$ \Rightarrow \sqrt 7 \,+ $$  $$\sqrt 6 \,- $$  $$\sqrt 6 \,- $$   $$\sqrt 5 \,+ $$  $$\sqrt 5 \,+ $$  $$\sqrt 4 \,- $$  $$\sqrt 8 \,- $$  $$\sqrt 7 \,+ $$  $$\sqrt 9 \,+ $$  $$\sqrt 8 $$
$$\eqalign{ & \Rightarrow \sqrt 4 + \sqrt 9 \cr & \Rightarrow 2 + 3 \cr & \Rightarrow 5 \cr} $$

Given expression,
$${\frac{1}{{1 + a + {b^{ - 1}}}} + }$$   $${\frac{1}{{1 + b + {c^{ - 1}}}} + }$$   $${\frac{1}{{1 + c + {a^{ - 1}}}}}$$
  $$ = \frac{1}{{1 + a + {b^{ - 1}}}} + $$   $$\frac{b^{ - 1}}{{{b^{ - 1}} + 1 + {b^{ - 1}}{c^{ - 1}}}} + $$    $$\frac{1}{{a + ac + 1}}$$
  $$ = \frac{1}{{1 + a + {b^{ - 1}}}} + $$   $$\frac{{{b^{ - 1}}}}{{1 + {b^{ - 1}} + a}} + $$   $$\frac{a}{{a + {b^{ - 1}} + 1}}$$
$$\eqalign{ & = \frac{{1 + a + {b^{ - 1}}}}{{1 + a + {b^{ - 1}}}} \cr & = 1 \cr} $$
$$\left[ {\because abc = 1 \Rightarrow {{\left( {bc} \right)}^{ - 1}} = a \Rightarrow {b^{ - 1}}{c^{ - 1}} = a,{\text{and }}ac = {b^{ - 1}}} \right]$$

$$\eqalign{ & {4^{61}} + {4^{62}} + {4^{63}} + {4^{64}} \cr & = {4^{61}}\left( {{4^0} + {4^1} + {4^2} + {4^3}} \right) \cr & = {4^{61}} \times 85 \cr} $$
Now check with option 85 is divisible by 17.

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

$$\sqrt {20} + \sqrt {12} + \root 3 \of {729} - \frac{4}{{\sqrt 5 - \sqrt 3 }} - \sqrt {81} $$
$$ = 2\sqrt 5 + 2\sqrt 3 + 9\, - $$   $$\left( {\frac{4}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}} \right)$$     $$\, - \,9$$
$$\eqalign{ & = 2\sqrt 5 + 2\sqrt 3 + 9 - \left( {\frac{{4\left( {\sqrt 5 + \sqrt 3 } \right)}}{2}} \right) - 9 \cr & = 2\sqrt 5 + 2\sqrt 3 + 9 - 2\sqrt 5 - 2\sqrt 3 - 9 \cr & = 0 \cr} $$

$$\eqalign{ & \sqrt {40 + \sqrt {9\sqrt {81} } } \cr & \Rightarrow \sqrt {40 + \sqrt {9 \times 9} } \cr & \Rightarrow \sqrt {40 + 9} \cr & \Rightarrow \sqrt {49} \cr & \Rightarrow 7 \cr} $$

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