$$\eqalign{ & \frac{{{{\left( {a + b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)}} \cr & = \frac{{\left( {a + b} \right)\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a - b} \right)}} \cr & = \frac{{a + b}}{{a - b}}\, \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {0.00005746} \cr & \Rightarrow \sqrt {\frac{{5746}}{{100000000}}} \cr & \Rightarrow \frac{{75.8}}{{10000}} \cr & \Rightarrow 0.00758 \cr} $$

$$\eqalign{ & \frac{{ab}}{{a + b}} = \frac{{\frac{x}{{x + y}} \times \frac{y}{{x - y}}}}{{\frac{x}{{x + y}} + \frac{y}{{x - y}}}} \cr & \frac{{ab}}{{a + b}} = \frac{{\frac{{xy}}{{{x^2} - {y^2}}}}}{{\frac{{x\left( {x - y} \right) + y\left( {x + y} \right)}}{{{x^2} - {y^2}}}}} \cr & \frac{{ab}}{{a + b}} = \frac{{xy}}{{{x^2} + {y^2}}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {\frac{{0.25}}{{0.0009}}} \times \sqrt {\frac{{0.09}}{{0.36}}\,\,} \cr & \Rightarrow \sqrt {\frac{{25}}{9} \times 100} \times \sqrt {\frac{9}{{36}}\,\,} \cr & \Rightarrow \frac{{5 \times 10}}{3} \times \frac{3}{6} \cr & \Rightarrow \frac{{25}}{3} \cr & \Rightarrow 8\frac{1}{3} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {32} {\text{ }} - \sqrt {128} {\text{ + }}\sqrt {50} \cr & \Rightarrow \sqrt {16 \times 2} - \sqrt {64 \times 2} + \sqrt {25 \times 2} \cr & \Rightarrow 4\sqrt 2 - 8\sqrt 2 + 5\sqrt 2 \cr & \Rightarrow \sqrt 2 \cr & \Rightarrow 1.414 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}} \div \frac{4}{7}\left( {\frac{2}{5} + \frac{3}{{10}}} \right){\text{of }}\frac{{\frac{1}{2} + \frac{1}{3}}}{{\frac{1}{2} - \frac{1}{3}}} \cr & = \frac{{\frac{3}{2}}}{{\frac{1}{2}}} \div \frac{4}{7}\left( {\frac{{20 + 15}}{{50}}} \right) \times \frac{{\frac{5}{6}}}{{\frac{1}{6}}} \cr & = \frac{3}{2} \times \frac{2}{1} \div \frac{4}{7} \times \frac{{35}}{{50}} \times \frac{5}{6} \times \frac{6}{1} \cr & = \frac{3}{2} \times \frac{2}{1} \div 2 \cr & = \frac{3}{2} \times 2 \div \frac{1}{2} \cr & = \frac{3}{2} \cr} $$

Given expression,
$$ = 2{a^3} - \left[ {3{a^3} + 4{b^3} - \left\{ {2{a^3} + \left( { - 7{a^3}} \right)} \right\}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right]$$
$$\eqalign{ & = 2{a^3} - \left[ {3{a^3} + 4{b^3} - \left\{ { - 5{a^3}} \right\}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right] \cr & = 2{a^3} - \left[ {3{a^3} + 4{b^3}{\text{ + 5}}{a^3}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right] \cr & = 2{a^3} - \left[ {13{a^3} - 3{b^3}} \right] \cr & = 2{a^3} - 13{a^3} + 3{b^3} \cr & = - 11{a^3} + 3{b^3} \cr} $$

$$\eqalign{ & \frac{a}{{b + c}} = k \cr & \Rightarrow a = k\left( {b + c} \right)\,....(i) \cr & \frac{b}{{c + a}} = k \cr & \Rightarrow b = k\left( {c + a} \right)\,....(ii) \cr & \frac{c}{{a + b}} = k \cr & \Rightarrow c = k\left( {a + b} \right)\,....(iii) \cr & {\text{Adding (i), (ii) and (iii) we get,}} \cr & a + b + c \cr & = k\left( {2a + 2b + 2c} \right) \cr & \Rightarrow k = \frac{{a + b + c}}{{2(a + b + c)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{Length of reel}} \cr & {\text{ = (32}} \times {\text{120 + 80)}}\,{\text{cm}} \cr & {\text{ = 3920}}\,{\text{cm}} \cr & {\text{ = 39}}{\text{.20}}\,{\text{m}} \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{1}{3}\,\, \Rightarrow b = 3a; \cr & \frac{b}{c} = 2\,\,\, \Rightarrow c = \frac{b}{2} = \frac{{3a}}{2}; \cr & \frac{c}{d} = \frac{1}{2}\,\,\, \Rightarrow d = 2c = 2\left( {\frac{{3a}}{2}} \right) = 3a; \cr & \frac{d}{e} = 3\,\,\, \Rightarrow e = \frac{d}{3} = \left( {\frac{{3a}}{3}} \right) = a; \cr & \frac{e}{f} = \frac{1}{4}\,\,\, \Rightarrow f = 4e = 4a \cr & \therefore \frac{{abc}}{{def}} = \frac{{\left( a \right)\left( {3a} \right)\left( {\frac{{3a}}{2}} \right)}}{{\left( {3a} \right)\left( a \right)\left( {4a} \right)}} \cr & \frac{{abc}}{{def}} = \frac{9}{2}{a^3} \times \frac{1}{{12{a^3}}} \cr & \frac{{abc}}{{def}} = \frac{3}{8} \cr} $$

$$\eqalign{ & \frac{m}{n}{\text{ = }}\frac{4}{3}{\text{ and }}\frac{r}{t}{\text{ = }}\frac{9}{{14}} \cr & \Rightarrow \frac{{mr}}{{nt}} = \frac{4}{3} \times \frac{9}{{14}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{7} \cr & \therefore \frac{{3mr - nt}}{{4nt - 7mr}}{\text{ }} \cr & = \frac{{3\frac{{mr}}{{nt}} - 1}}{{4 - 7\frac{{mr}}{{nt}}}} \cr & = \frac{{3 \times \frac{6}{7} - 1}}{{4 - 7 \times \frac{6}{7}}} \cr & = \frac{{\frac{{18}}{7} - 1}}{{4 - 6}} \cr & = \frac{{11}}{7} \times \left( { - \frac{1}{2}} \right) \cr & = - \frac{{11}}{{14}} \cr} $$

$$\eqalign{ & a + b + c = 0 \cr & \Rightarrow a = - \left( {b - c} \right) \cr & \Rightarrow {a^2} = {\left( {b + c} \right)^2} \cr & \therefore \frac{{{a^2}}}{{\left( {{a^2} - bc} \right)}} + \frac{{{b^2}}}{{\left( {{b^2} - ca} \right)}} + \frac{{{c^2}}}{{\left( {{c^2} - ab} \right)}} \cr} $$
$$ = \frac{{{{\left( {b + c} \right)}^2}}}{{{{(b + c)}^2} - bc}} + \frac{{{b^2}}}{{{b^2} + c\left( {b + c} \right)}} + \frac{{{c^2}}}{{{c^2} + b\left( {b + c} \right)}}$$
$$ = \frac{{{{\left( {b + c} \right)}^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{b^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{c^2}}}{{{b^2} + {c^2} + bc}}$$
$$\eqalign{ & = \frac{{{b^2} + {c^2} + 2bc + {b^2} + {c^2}}}{{{b^2} + {c^2} + bc}} \cr & = \frac{{2\left( {{b^2} + {c^2} + bc} \right)}}{{{b^2} + {c^2} + bc}} \cr & = 2 \cr} $$

$$\eqalign{ & \frac{{{a^2} - {b^2} - 2bc - {c^2}}}{{{a^2} + {b^2} + 2ab - {c^2}}}\, \cr & = \frac{{{a^2} - ({b^2} + 2bc + {c^2})}}{{({a^2} + {b^2} + 2ab) - {c^2}}} \cr & = \frac{{{a^2} - {{(b + c)}^2}}}{{{{(a + b)}^2} - {c^2}}} \cr & = \frac{{\left( {a + b + c} \right)\left( {a - b - c} \right)}}{{\left( {a + b + c} \right)\left( {a + b - c} \right)}} \cr & = \frac{{a - b - c}}{{a + b - c}} \cr} $$

$$\eqalign{ & {\left( {s - a} \right)^2} + {\left( {s - b} \right)^2} + {\left( {s - c} \right)^2} + {s^2} \cr & = \left( {{s^2} + {a^2} - 2sa} \right) + \left( {{s^2} + {b^2} - 2sb} \right) + \cr & \,\,\,\,\,\,\left( {{s^2} + {c^2} - 2sc} \right) + {s^2} \cr & = 4{s^2} + ({a^2} + {b^2} + {c^2}) - 2s(a + b + c) \cr & = {(2s)^2} + ({a^2} + {b^2} + {c^2}) - \cr & \,\,\,\,\,\,(a + b + c)(a + b + c) \cr & = {(a + b + c)^2} + \left( {{a^2} + {b^2} + {c^2}} \right) - \cr & \,\,\,\,\,\,{(a + b + c)^2} \cr & = {a^2} + {b^2} + {c^2} \cr} $$

$$\left[ {1 + \frac{1}{{x + 1}}} \right]$$ $$\left[ {1 + \frac{1}{{x + 2}}} \right]$$ $$\,\left[ {1 + \frac{1}{{x + 3}}} \right]$$ $$\left[ {1 + \frac{1}{{x + 4}}} \right]$$
$$ = \left[ {\frac{{\left( {x + 1} \right) + 1}}{{x + 1}}} \right]$$  $$\left[ {\frac{{\left( {x + 2} \right) + 1}}{{x + 2}}} \right]$$ $$\left[ {\frac{{\left( {x + 3} \right) + 1}}{{x + 3}}} \right]$$ $$\left[ {\frac{{\left( {x + 4} \right) + 1}}{{x + 4}}} \right]$$
$$\eqalign{ & = \left( {\frac{{x + 2}}{{x + 1}}} \right)\left( {\frac{{x + 3}}{{x + 2}}} \right)\left( {\frac{{x + 4}}{{x + 3}}} \right)\left( {\frac{{x + 5}}{{x + 4}}} \right) \cr & = \frac{{x + 5}}{{x + 1}} \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{4}{5}{\text{ and }}\frac{b}{c} = \frac{{15}}{{16}} \cr & \Rightarrow \left( {\frac{a}{b} \times \frac{b}{c}} \right) = \left( {\frac{4}{5} \times \frac{{15}}{{16}}} \right) \cr & \Rightarrow \frac{a}{c} = \frac{3}{4} \cr & \therefore \,\frac{{{c^2} - {a^2}}}{{{c^2} + {a^2}}} \cr & = \frac{{1 - \left( {\frac{{{a^2}}}{{{c^2}}}} \right)}}{{1 + \left( {\frac{{{a^2}}}{{{c^2}}}} \right)}} \cr & = \frac{{1 - {{\left( {\frac{a}{c}} \right)}^2}}}{{1 + {{\left( {\frac{a}{c}} \right)}^2}}} \cr & = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}} \cr & = \frac{{\left( {\frac{7}{{16}}} \right)}}{{\left( {\frac{{25}}{{16}}} \right)}} \cr & = \frac{7}{{25}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{1}{8} + \frac{1}{{{8^2}}} + \frac{1}{{{8^3}}} + \frac{1}{{{8^4}}} + \frac{1}{{{8^5}}} \cr & \Rightarrow \frac{1}{8} + \frac{1}{{64}} + \frac{1}{{512}} + \frac{1}{{4096}} + \frac{1}{{32768}} \cr} $$
  ⇒ 0.125 + 0.015625 + 0.00195313 + 0.00024414 + 0.0000305175
  ⇒ 0.143

According to question,
$$\sqrt {{{\left( {0.798} \right)}^2} + 0.404 \times 0.798 + {{\left( {0.202} \right)}^2}} + 1\,$$
$$ \Rightarrow \sqrt {{{\left( {0.798} \right)}^2} + {{\left( {0.202} \right)}^2} + 2 \times 0.202 \times 0.798\,} + 1\,$$
$$ \Rightarrow \sqrt {{{\left( {0.798 + 0.202} \right)}^2}} + 1\,\,\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + {b^2} + 2ab} \right]$$
$$\eqalign{ & \Rightarrow \sqrt {{{\left( 1 \right)}^2}} + 1 \cr & \Rightarrow 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}} \cr & {\text{let }}a = 0.73{\text{ and }}b = 0.27 \cr} $$
 $$ = \frac{{\left( {0.73 + 0.27} \right)\left[ {{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27} \right]}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}}$$
$$\eqalign{ & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = 0.73 + 0.27 \cr & = 1 \cr} $$

$$\eqalign{ & \frac{{{{\left( {a - b} \right)}^2} - {{\left( {a + b} \right)}^2}}}{{ - 4a}}{\text{ = }}\frac{x}{y} \cr & \Rightarrow \frac{{\left( {{a^2} + {b^2} - 2ab} \right) - \left( {{a^2} + {b^2} + 2ab} \right)}}{{ - 4a}} = \frac{x}{y} \cr & \Rightarrow \frac{{ - 4ab}}{{ - 4a}} = \frac{x}{y} \cr & \Rightarrow b = \frac{x}{y} \cr & \Rightarrow x = yb \cr} $$

4376 + 3209 - 1784 + 97 = 3125 + x
⇒ 7682 - 1784 = 3125 + x
or, x = 7682 - 1784 - 3125
⇒ x = 2773
Hence,
the number is 2773

Let the missing number is a
Given,
$$\eqalign{ & 14 \times 627 \div \sqrt {\left( {1089} \right)} = {\left( a \right)^3} + 141 \cr & \frac{{14 \times 627}}{{\sqrt {1089} }} = {\left( a \right)^3} + 141 \cr & \frac{{14 \times 627}}{{33}} = {a^3} + 141 \cr & 266 = {a^3} + 141 \cr & {a^3} = 125 \cr & a = \root 3 \of {125} \cr & a = 5 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{6}{{5 - \frac{5}{3}}} \div \frac{{4 - \frac{2}{{4 - \frac{1}{2}}}}}{{5 - \frac{3}{2}}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{2}{3} \times \frac{1}{2}} \right\}{\text{ }} \cr & = \frac{6}{{\left( {\frac{{10}}{3}} \right)}} \div \frac{{4 - \frac{2}{{\left( {\frac{7}{2}} \right)}}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{1}{3}} \right\} \cr & = \frac{{6 \times 3}}{{10}} \div \frac{{4 - \frac{{2 \times 2}}{7}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \times 1 \cr & = \frac{9}{5} \div \frac{{\left( {4 - \frac{4}{7}} \right)}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \cr & = \frac{9}{5} \div \left( {\frac{{24}}{7} \times \frac{2}{7}} \right) - \frac{2}{5} \cr & = \frac{9}{5} \times \frac{{49}}{{48}} - \frac{2}{5} \cr & = \frac{{147}}{{80}} - \frac{2}{5} \cr & = \frac{{147 - 32}}{{80}} \cr & = \frac{{115}}{{80}} \cr & = \frac{{23}}{{16}} \cr & = 1\frac{7}{{16}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & {\text{I = }}\frac{3}{5} \div \frac{5}{6} \cr & {\text{I = }}\frac{3}{5} \times \frac{6}{5} = \frac{18}{{25}} \cr & \cr & {\text{II = 3}} \div \left[ {\frac{4}{5} \times \frac{1}{6}} \right] \cr & {\text{II = 3}} \times \frac{{30}}{4} \cr & {\text{II = }}\frac{{45}}{2} \cr & \cr & {\text{III = }}\left[ {3 \div \left( {4 \div 5} \right)} \right] \div 6\, \cr & {\text{III = }}\left[ {3 \times \frac{5}{4}} \right] \times \frac{1}{6}\, \cr & {\text{III = }}\frac{5}{8} \cr & \cr & {\text{IV = 3}} \div {\text{4}}\left( {5 \div 6} \right) \cr & {\text{IV = 3}} \div {\text{4}} \times \frac{5}{6} \cr & {\text{IV = 3}} \times \frac{6}{{20}} \cr & {\text{IV = }}\frac{9}{{10}} \cr

$$\eqalign{ & \frac{{x + 1}}{{x - 1}}{\text{ = }}\frac{a}{b}{\text{ }} \cr & {\text{By componendo and dividendo}} \cr & \frac{{x + 1 + x - 1}}{{x + 1 - x + 1}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{{2x}}{2} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow x = \frac{{a + b}}{{a - b}}.....(i) \cr & {\text{Again ,}}\frac{{1 - y}}{{1 + y}}{\text{ = }}\frac{b}{a}{\text{ }} \cr & \Rightarrow \frac{{1 + y}}{{1 - y}} = \frac{a}{b} \cr & \Rightarrow \frac{{1 + y + 1 - y}}{{1 + y - 1 + y}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{2}{{2y}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{1}{y} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow y = \frac{{a - b}}{{a + b}}.....(ii) \cr & {\text{Subtracting equation (ii) from (i) we get }} \cr & \therefore x - y = \frac{{a + b}}{{a - b}} - \frac{{a - b}}{{a + b}} \cr & \left[ {\because {{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2} = 4ab\,\,\,and\,\,{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right] \cr & = \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {a - b} \right)\left( {a + b} \right)}} \cr & = \frac{{4ab}}{{{a^2} - {b^2}}} \cr & {\text{Multiply equation (i) and (ii) we get}} \cr & xy = \frac{{a + b}}{{a - b}} \times \frac{{a - b}}{{a + b}} = 1 \cr & \therefore {\text{Expression,}} \cr & {\text{ = }}\frac{{x - y}}{{1 + xy}} = \frac{{\frac{{4ab}}{{{a^2} - {b^2}}}}}{{1 + 1}} \cr &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4ab}}{{2\left( {{a^2} - {b^2}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2ab}}{{{a^2} - {b^2}}} \cr} $$

200 ÷ 25 × 4 + 12 - 3
= $$\frac{{200}}{{25}}$$ × 4 + 12 - 3
= 8 × 4 + 12 - 3
= 32 + 12 - 3
= 44 - 3
= 41

$$\eqalign{ & {\text{Given,}} \cr & \left( {x + \frac{1}{x}} \right) = 3{\text{ }} \cr & {\text{On squaring both sides, we get}} \cr & {\left( {x + \frac{1}{x}} \right)^2} = {3^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr} $$

As we know
a³ + b³ + c³ + 3abc = (a² + b² + c² - ab - bc - ca)(a + b + c)
When, a + b + c = 0
Then a³ + b³ + c³ - 3abc = 0
When, x + y + z = 0
⇒ x³ + y³ + z³ = 3xyz
⇒ x³ + y³ + z³ + 3xyz = 3xyz + 3xyz
⇒ x³ + y³ + z³ + 3xyz = 6xyz

$$\eqalign{ & {\text{ = }}\, - 7m - \left[ {3n - \left\{ {8m - \left( {4n - 10m} \right)} \right\}} \right]\,\,\, \cr & = \, - 7m - \left[ {3n - \left\{ {8m - 4n + 10m} \right\}} \right]\, \cr & = - 7m - \left[ {3n - 18m + 4n} \right] \cr & = - 7m - \left[ {7n - 18m} \right] \cr & = - 7m - 7n + 18m\,\,\, \cr & = \,11m - 7n\,\, \cr} $$

$$\eqalign{ & \frac{2}{a} + \frac{1}{b} \cr & = \frac{{2b + a}}{{ab}} \cr & = \frac{6}{4} \cr & = \frac{3}{2} \cr} $$