L.C.M. of 3, 4, 5, 6, 8 is 120
Now 120 = 2 × 2 × 2 × 3 × 5
To make it a perfect square, it must be multiplied by 2 × 3 × 5
So, required number
$$\eqalign{ & = {2^2} \times {2^2} \times {3^2} \times {5^2} \cr & = 3600 \cr} $$

294 = 7 × 7 × 2 × 3
To make it a perfect square, it must be multiplied by 2 × 3 i.e.,6
∴ Required number = 6

$$\eqalign{ & = \sqrt {\frac{3}{{11}}} \cr & = \sqrt {\frac{{3 \times 11}}{{11 \times 11}}} \cr & = \frac{{\sqrt {33} }}{{11}} \cr & = \frac{{5.745}}{{11}} \cr & = 0.5223 \cr} $$

$$\eqalign{ & a = \frac{{\sqrt 3 }}{2}{\text{ (given)}} \cr & \therefore \sqrt {1 + a} + \sqrt {1 - a} \cr & = \sqrt {1 + \frac{{\sqrt 3 }}{2}} + \sqrt {1 - \frac{{\sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2 + \sqrt 3 }}{2}} + \sqrt {\frac{{2 - \sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2\left( {2 + \sqrt 3 } \right)}}{4}} + \sqrt {\frac{{2\left( {2 - \sqrt 3 } \right)}}{4}} \cr & = \sqrt {\frac{{4 + 2\sqrt 3 }}{4}} + \sqrt {\frac{{4 - 2\sqrt 3 }}{4}} \cr} $$
$$ = \sqrt {\frac{{3 + 1 + 2 \times \sqrt 3 \times 1}}{2}} + $$     $$\sqrt {\frac{{3 + 1 - 2 \times \sqrt 3 \times 1}}{2}} $$     \[\because \left\{ \begin{gathered} {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} - 2.\sqrt 3 .1 = {\left( {\sqrt 3 - 1} \right)^2} \hfill \\ {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} + 2.\sqrt 3 .1 = {\left( {\sqrt 3 + 1} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a + b} \right)^2} \hfill \\ \end{gathered} \right\}\]
$$\eqalign{ & = \sqrt {\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{2}} + \sqrt {\frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}} \cr & = \frac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{2} \cr & = \frac{{2\sqrt 3 }}{2} \cr & = \sqrt 3 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 2\sqrt {20} - \sqrt {32} + \sqrt {50} }}{\text{ }} \cr & = \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 2 \times 2\sqrt 5 - 2 \times 2\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 4\sqrt 5 - 4\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{5 + \sqrt {10} }}{{\sqrt 5 + \sqrt 2 }} \cr & = \frac{{\sqrt 5 \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\sqrt 5 + \sqrt 2 }} \cr & = \sqrt 5 \cr} $$

$$\eqalign{ & {\text{Required number to be added}} \cr & = {9^3} - 710 \cr & = 729 - 710 \cr & = 19 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & = \root 4 \of {{{\left( {625} \right)}^3}} \cr & = {\left( {625} \right)^{\frac{3}{4}}} \cr & = {\left( {5 \times 5 \times 5 \times 5} \right)^{\frac{3}{4}}} \cr & = {\left( {{5^4}} \right)^{\frac{3}{4}}} \cr & = {5^3} \cr & = 125 \cr} $$

$$\frac{{{{\left( {0.75} \right)}^3}}}{{1 - 0.75}}$$ $${\text{ + }}$$$$\left[ {0.75 + {{\left( {0.75} \right)}^2} + 1} \right]$$
$$ = \frac{{{{\left( {0.75} \right)}^2} \times 0.75}}{{0.25}}$$   $${\text{ + }}$$ $$\left[ {0.75 + 0.5625 + 1} \right]$$
$$ = 0.5625 \times 3 \,\, + $$   $$\left[ {0.75 + 0.5625 + 1} \right]$$
$$\eqalign{ & = 1.6875 + 2.3125 \cr & = 4 \cr} $$
Square root of 4 = 2

$$\eqalign{ & \Leftrightarrow \sqrt y = 4x \cr & \Leftrightarrow y = {\left( {4x} \right)^2} \cr & \Leftrightarrow y = 16{x^2} \cr & \Leftrightarrow \frac{{{x^2}}}{y} = \frac{1}{{16}} \cr} $$

$$\eqalign{ & {\text{We have to find}} \cr & = \sqrt {9{x^2} + 25 - 30x} \cr} $$
 $$ = \sqrt {{{\left( {3x} \right)}^2} - 2.3x.5 + {{\left( { - 5} \right)}^2}} $$   $$\,\,\,\,\,\,\,\left\{ {\because {a^2} - 2ab + {b^2} = {{\left( {a + b} \right)}^2}} \right\}$$
$$\eqalign{ & = \sqrt {{{\left( {3x - 5} \right)}^2}} \cr & = 3x - 5 \cr} $$

$$\eqalign{ & = \root 3 \of {\frac{{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}}{{0.4 \times 0.4 \times 0.4 + 0.08 \times 0.08 \times 0.08}}} \cr & = \root 3 \of {\frac{{0.008 + 0.000064}}{{0.064 + 0.000512}}} \cr & = \root 3 \of {\frac{{0.008064}}{{0.064512}}} \cr & = \root 3 \of {\frac{{8064}}{{64512}}} \cr & = \root 3 \of {\frac{1}{8}} \cr & = \frac{1}{2} \cr & = 0.5 \cr} $$

$$\eqalign{ & {\text{Clearly, }}a*b = \sqrt {{a^2} + {b^2}} \cr & \therefore 5*12 \cr & = \sqrt {{5^2} + {{12}^2}} \cr & = \sqrt {25 + 144} \cr & = \sqrt {169} \cr & = 13 \cr} $$

$$\eqalign{ & \Rightarrow \frac{1}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)}} \cr & = \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} \cr & = \frac{{\left( {2.2361 + 1.7321} \right)}}{2} \cr & = \frac{{3.9682}}{2} \cr & = 1.9841{\text{ }} \cr} $$

$$\eqalign{ & \,\,\,\,\,\,1|\overline 1 \,\,\overline {10} \,\,\overline {25} \,(105 \cr & \,\,\,\,\,\,\,\,\,|\,\,1 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,20|\,\,\,\,\,\,10 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,0 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & 205|\,\,\,\,\,\,\,10\,25 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,10\,25 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{X}} \cr & {\text{Number of rows}} \cr & {\text{ = }}\sqrt {10914 + 111} \cr & = \sqrt {11025} \cr & = 105 \cr} $$

$$\eqalign{ & = \sqrt {0.000064} \cr & = \sqrt {\frac{{64}}{{{{10}^6}}}} \cr & = \frac{8}{{{{10}^3}}} \cr & = \frac{8}{{1000}} \cr & = 0.008 \cr & \therefore \root 3 \of {\sqrt {0.000064} } \cr & = \root 3 \of {0.008} \cr & = \root 3 \of {\frac{8}{{1000}}} \cr & = \frac{2}{{10}} \cr & = 0.2 \cr} $$

Let the number of girls in the group be x
Then, number of oranges given to each girl = 2x
$$\eqalign{ & \therefore x \times 2x = 1250 \cr & \Leftrightarrow 2{x^2} = 1250 \cr & \Leftrightarrow {x^2} = 625 \cr & \Leftrightarrow x = 25 \cr} $$

Given expression,
$$ = \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$     $$ + \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$   $$ \times \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}}$$   $$ + \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}}$$   $$ \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}}$$
$$ = \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}} + \frac{{{{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}}$$     $$ + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {3 - 1} \right)}}$$
$$ = \left[ {{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}} \right]$$     $$ + \frac{{4 - 2\sqrt 3 }}{2}$$
$$ = 2\left( {4 + 3} \right) + 2 - \sqrt 3 $$
$$ = 16 - \sqrt 3 $$

$$\eqalign{ & \because x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = {\left( {3 + \sqrt 8 } \right)^2} \cr & \Rightarrow {x^2} = {3^2} + {\left( {\sqrt 8 } \right)^2} + 2 \times 3 \times \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 6\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr} $$
$$\therefore {x^2} + \frac{1}{{{x^2}}}$$
$$ = \left( {17 + 12\sqrt 2 } \right)$$   $$ + \frac{1}{{\left( {17 + 12\sqrt 2 } \right)}}$$   $$ \times \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{\left( {17 - 12\sqrt 2 } \right)}}$$
$$\eqalign{ & = \left( {17 + 12\sqrt 2 } \right) + \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{289 - 288}} \cr & = \left( {17 + 12\sqrt 2 } \right) + \left( {17 - 12\sqrt 2 } \right) \cr & = 17 + 17 \cr & = 34 \cr} $$

$$\eqalign{ & 0.000326 = \frac{{326}}{{{{10}^6}}} \cr & \,\,\,\,1|\overline 3 \,\,\overline {26} \,\,(18 \cr & \,\,\,\,\,\,\,|\,\,1 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & 28|\,\,\,2\,26 \cr & \,\,\,\,\,\,\,|\,\,\,2\,24 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr} $$
∴ Required number to be subtracted
$$\eqalign{ & = \frac{2}{{{{10}^6}}} \cr & = 0.000002 \cr} $$

$$\eqalign{ & = \sqrt {0.01} \times \root 3 \of {0.008} - 0.02{\text{ }} \cr & = \sqrt {{{\left( {0.1} \right)}^2}} \times \root 3 \of {{{\left( {0.2} \right)}^3}} - 0.02 \cr & = 0.1 \times 0.2 - 0.02 \cr & = 0.02 - 0.02 \cr & = 0 \cr} $$

Given expression,
$$ = \frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \times \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 + \sqrt 8 } \right)}}$$     $$ - \frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 + \sqrt 7 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {\sqrt 7 + \sqrt 6 } \right)}}$$   $$ - \frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 + \sqrt 5 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {\sqrt 5 + \sqrt 4 } \right)}}$$
$$ = \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {9 - 8} \right)}} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {8 - 7} \right)}}$$     $$ + \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {7 - 6} \right)}}$$   $$ - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {6 - 5} \right)}}$$   $$ + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {5 - 4} \right)}}$$
$$ = \left( {\sqrt 9 + \sqrt 8 } \right) - \left( {\sqrt 8 + \sqrt 7 } \right)$$     $$ + \left( {\sqrt 7 + \sqrt 6 } \right)$$   $$ - \left( {\sqrt 6 + \sqrt 5 } \right)$$   $$ + \left( {\sqrt 5 + \sqrt 4 } \right)$$
$$ = \left( {\sqrt 9 + \sqrt 4 } \right)$$
$$ = 3 + 2$$
$$ = 5$$

Given expressing,
$$ = \frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}\frac{1}{{\sqrt 2 + \sqrt 3 }}$$     $$ + \frac{1}{{\sqrt 3 + \sqrt 4 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }}{\text{ + }}\frac{1}{{\sqrt 3 + \sqrt 2 }}$$     $$ + \frac{1}{{\sqrt 4 + \sqrt 3 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }} \times $$   $$\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}{\text{ + }}$$   $$\frac{1}{{\sqrt 3 + \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + $$   $$\frac{1}{{\sqrt 4 + \sqrt 3 }} \times $$   $$\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} + $$   $$...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }} \times $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{\sqrt {121} - \sqrt {120} }}$$
$$ = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}$$     $$ + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}}$$   $$ + ...... + $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{121 - 120}}$$
$$ = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 $$     $$ + \sqrt 4 - \sqrt 3 $$   $$ + ...... + $$   $$\sqrt {121} - \sqrt {120} $$
$$ = - 1 + \sqrt {121} $$
$$ = - 1 + 11$$
$$ = 10$$

Clearly, the required number must be a perfect square. Since a number having 8 as the unit's digit cannot be a perfect square, so 121108 is not a perfect square.

$$\eqalign{ & = 262144 \cr & = 8 \times 8 \times 8 \times 8 \times 8 \times 8 \cr & = {8^6} \cr & \therefore \root 3 \of {262144} \cr & = {8^2} \cr & = 64 \cr & {\text{Let, }} \cr & {\text{1728}} \div \root 3 \of {262144} \times x - 288 = 4491 \cr & {\text{Then,}} \cr & \Leftrightarrow 1728 \div 64 \times x - 288 = 4491 \cr & \Leftrightarrow 27x = 4779 \cr & \Leftrightarrow x = \frac{{4779}}{{27}} \cr & \Leftrightarrow x = 177 \cr} $$

$$\eqalign{ & 8|262144 \cr & - - - - - - - - \cr & 8|\,\,\,32768 \cr & - - - - - - - - \cr & 8|\,\,\,\,4096 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,512 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,64 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,\,8 \cr & - - - - - - - - \cr} $$

$$\eqalign{ & \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr & = 5 + 2\sqrt 6 \cr & \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr & = 5 - 2\sqrt 6 \cr & \therefore {\text{ }}{a^2} + {b^2} \cr & = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr & = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr & = 2\left( {25 + 24} \right) \cr & = 2 \times 49 \cr & = 98 \cr} $$

$$\eqalign{ & {\text{ }}a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & \,\,\,\,\,\, {\text{ = }}\frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \times \frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 + 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 + \sqrt 5 }}{2}} \right) \cr & {\text{ }}b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \,\,\,\,\,\,\, = \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \times \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 - 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 - \sqrt 5 }}{2}} \right) \cr & \,\,\,\,\,\,\, \therefore {a^2} + {b^2} \cr & \,\,\,\,\,\,\, = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} + {\left( {\frac{{3 - \sqrt 5 }}{2}} \right)^2} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2}}}{4} + \frac{{{{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2} + {{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{9 + 2.3.\sqrt 5 + 5 + 9 - 2.3.\sqrt 5 + 5}}{4} \cr & \,\,\,\,\,\,\, = \frac{{2\left( {9 + 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = \frac{{28}}{4} \cr & \,\,\,\,\,\,\, = 7 \cr & {\text{Also, }} \cr & ab = \frac{{\left( {3 + \sqrt 5 } \right)}}{2} \times \frac{{\left( {3 - \sqrt 5 } \right)}}{2} \cr & \,\,\,\,\,\,\, = \frac{{\left( {9 - 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = 1 \cr & \,\,\,\,\,\,\, \therefore \left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right) \cr & \,\,\,\,\,\,\, = \frac{{\left( {{a^2} + {b^2}} \right) + ab}}{{\left( {{a^2} + {b^2}} \right) - ab}} \cr & \,\,\,\,\,\,\, = \frac{{7 + 1}}{{7 - 1}} \cr & \,\,\,\,\,\,\, = \frac{8}{6} \cr & \,\,\,\,\,\,\, = \frac{4}{3} \cr} $$

$$\eqalign{ & \,\,\,\,\,8|\overline {77} \,\,\overline {00} \,\,(87 \cr & \,\,\,\,\,\,\,\,\,|\,\,64 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & 167|\,\,\,13\,00 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,11\,69 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,31 \cr & \therefore {\text{Number to be added}} \cr & = {\left( {88} \right)^2} - 7700 \cr & = 7744 - 7700 \cr & = 44 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,8|\overline {68} \,\,\overline {06} \,\,\overline {21} \,(824 \cr & \,\,\,\,\,\,\,\,\,\,\,\,|\,\,64 \cr & \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,162|\,\,\,\,4\,06 \cr & \,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,3\,24 \cr & \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 1644|\,\,\,\,\,\,\,\,\,\,82\,21 \cr & \,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,65\,76 \cr & \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,45 \cr & \therefore {\text{Number to be added}} \cr & = {\left( {825} \right)^2} - 680621 \cr & = 680625 - 680621 \cr & = 4 \cr} $$

$$\eqalign{ & = \frac{{1 + \sqrt {0.01} }}{{1 - \sqrt {0.1} }} \cr & = \frac{{1 + 0.1}}{{1 - 0.316}} \cr & = \frac{{1.1}}{{0.684}} \cr & = \frac{{1100}}{{684}} \cr & = 1.6 \cr & \,\,\,\,\,\,\,3|\overline 0 \,.\,\overline {10} \,\,\overline {00} \,\,\overline {00} \,(0.316 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,61|\,\,\,\,\,\,\,\,\,\,100 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,61 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 626\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,3900 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,3756 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr} $$

$$\eqalign{ & = \sqrt {{\text{535}}{\text{.9225}}} \cr & = \sqrt {\frac{{{\text{5359225}}}}{{10000}}} \cr & = \frac{{2315}}{{100}} \cr & = 23.15 \cr & \,\,\,\,\,\,\,\,\,2|\overline 5 \,\,\overline {35} \,\,\overline {92} \,\,\overline {25} \,\,(2315 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,\,43|\,\,1\,35 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,1\,2\,9 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,461\,|\,\,\,\,\,\,\,\,\,6\,92 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,4\,61 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & 4625\,|\,\,\,\,\,\,\,\,\,\,\,\,231\,25 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,231\,25 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x}} \cr} $$