$$\eqalign{ & {\text{Let the number be }}x \cr & {\text{Given,}} \cr & \frac{{\left( {923 - 347} \right)}}{x} = 32{\text{ }} \cr & \Rightarrow x = \frac{{576}}{{32}} = 18 \cr & \Rightarrow x = 18 \cr} $$

Let the number be a
Given,
1559.95 - 7.99 × 24.96 - a² = 1154
⇒ 1560 - 8 × 25 - a² = 1154
⇒ 1360 - 200 - a² = 1154
⇒ a² = 1360 - 1154
⇒ a² = 206
⇒ a ≈ $$\sqrt {206} $$
⇒ a ≈ 14

67 × 119 + 17 - 27 ÷ 259
⇒ 67 + 119 ÷ 17 × 27 - 259
⇒ 67 + 7 × 27 - 259
⇒ 67 + 189 - 259
⇒ 256 - 259
⇒ -3

$$\eqalign{ & {\text{Given}}, \cr & \left\{ {\left( {64 - 38} \right) \times 4} \right\} \div 13 \cr & = \frac{{26 \times 4}}{{13}} \cr & = 8 \cr} $$

$$\eqalign{ & {\text{Given, }} \cr & x + y = 2a \cr & \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr & = \frac{{a\left( {y - a} \right) + a\left( {x - a} \right)}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{ay - {a^2} + ax - {a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a\left( {x + y} \right) - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a.2a - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{2{a^2} - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = 0 \cr} $$

$$\eqalign{ & \left( {4{b^2} + \frac{1}{{{b^2}}}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - {\text{4 = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow \left( {2b + \frac{1}{b}} \right) = \sqrt 6 \cr & \left( {{\text{Cubeing the both sides}}} \right) \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3\times2b\times\frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) = 0 \cr} $$

Let the missing number is x
$$\eqalign{ & {\text{Given,}} \cr & \frac{{20 + 8 \times 0.5}}{{20 - x}}{\text{ = 12}} \cr & \Rightarrow \frac{{20 + 4}}{{20 - x}} = 12 \cr & \Rightarrow \frac{{24}}{{20 - x}} = 12 \cr & \Rightarrow 20 - x = \frac{{24}}{{12}} = 2 \cr & \Rightarrow x = 20 - 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 18 \cr & {\text{Hence, the number is 18}} \cr} $$

According to question,
$$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}}$$
As we know that
$${a^3} + {b^3} + {c^3} - 3abc = $$     $$\left( {a + b + c} \right)$$   $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$ = \frac{{{{(0.9)}^3} + {{(0.2)}^3} + {{(0.3)}^3} - 3 \times 0.9 \times 0.2 \times 0.3}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$
$$ = \frac{{\left( {0.9 + 0.2 + 0.3} \right)\left[ {{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9} \right]}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$
$$\eqalign{ & = 0.9 + 0.2 + 0.3 \cr & = 1.4 \cr} $$

$$\eqalign{ & {\text{Given, }} \cr & {\text{1}}\frac{4}{5} + 20 - 280 \div 25 \cr & = \frac{9}{5} + 20 - \frac{{280}}{{25}} \cr & = \frac{9}{5} + 20 - \frac{{56}}{5} \cr & = \frac{{9 + 100 - 56}}{5} \cr & = \frac{{53}}{5} \cr & = 10\frac{3}{5} \cr} $$

$$\eqalign{ & {\text{Let the number be }}x \cr & {\text{Given }} \cr & 421 \div 35 \times 299.97 \div 25.05 = {x^2} \cr & \approx {\left( x \right)^2} = 420 \div 35 \times 300 \div 25 \cr & \approx 12 \times 300 \div 25 = 12 \times 12 \cr & \approx {x^2} = 12 \times 12 \cr & \therefore x = \sqrt {12 \times 12} = 12 \cr & {\text{Hence the number is 12}} \cr} $$

$$\,\,\,\,\,\,\root 3 \of { - 2197} \times \root 3 \of { - 125} \div \root 3 \of {\frac{{27}}{{512}}} \,$$
  $$ = \root 3 \of {13 \times 13 \times 13 \times - 1} \,\, \times $$     $$\sqrt {5 \times 5 \times 5 \times - 1} \,\,\div $$     $$\root 3 \of {\frac{{27}}{{512}}} $$
$$\eqalign{ & = \left( {13 \times 5\root 3 \of { - 1 \times - 1} } \right) \div \frac{3}{8} \cr & = 13 \times 5 \times \frac{8}{3} \cr & = \frac{{520}}{3} \cr} $$

$$\eqalign{ & \left( {a + \frac{1}{a}} \right) = 6 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^2} = {6^2} \cr & \,\,\,\,\,\,\,\, = 36 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 36 \cr & \Rightarrow \left( {{a^2} + \frac{1}{{{a^2}}}} \right) = 34 \cr & \Rightarrow {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = {\left( {34} \right)^2} \cr & \Rightarrow {a^4} + \frac{1}{{{a^4}}} + 2 = 1156 \cr & \Rightarrow \left( {{a^4} + \frac{1}{{{a^4}}}} \right) = 1154 \cr} $$

$$\eqalign{ & \left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = }}{\left( {\sqrt {21} } \right)^2} = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 23 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 25 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{5}}^2} \cr & \Rightarrow x + \frac{1}{x} = 5 \cr & \therefore \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right) \cr & = 23 \times 5 \cr & = 115 \cr} $$

$$\eqalign{ & 0 < x < 1, \cr & {\text{Let }}x = \frac{4}{{10}} \cr & {\text{So}},\sqrt x = \frac{2}{{\sqrt {10} }}\,\& \, \cr & \,\,{x^2} = \frac{{16}}{{100}} = 0.16 \cr & {\text{Now}}, \cr & \because 0.16 < \frac{4}{{10}} < \frac{2}{{\sqrt {10} }} \cr & \therefore {x^2} < x < \sqrt x \cr} $$

$$\eqalign{ & {\text{Given,}}\frac{a}{b}{\text{ + }}\frac{b}{a}{\text{ = 2}} \cr & \Rightarrow \frac{{{a^2} + {b^2}}}{{ab}} = 2 \cr & \Rightarrow {a^2} + {b^2} = 2ab \cr & \Rightarrow {a^2} + {b^2} - 2ab = 0 \cr & \Rightarrow {\left( {a - b} \right)^2} = 0 \cr & \Rightarrow \left( {a - b} \right){\text{ = 0 }} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{\left( {x + 1} \right) - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right] - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left( {\frac{2}{{{x^2} - 1}} - \frac{2}{{{x^2} + 1}}} \right) - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{2\left( {{x^2} + 1} \right) - 2\left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}} \right] - \frac{4}{{{x^4} + 1}} \cr & = \frac{4}{{{x^4} - 1}} - \frac{4}{{{x^4} + 1}} \cr & = \frac{{4\left( {{x^4} + 1} \right) - 4\left( {{x^4} - 1} \right)}}{{\left( {{x^4} - 1} \right)\left( {{x^4} + 1} \right)}} \cr & = \frac{8}{{{x^8} - 1}} \cr} $$

Given expression,
$$\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]$$     $$\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + x.\frac{1}{x}} \right)} \right]$$
$$\eqalign{ & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = {x^6} - \frac{1}{{{x^6}}} \cr} $$

Given Expression
128.43 + 30.21 + ? = 173
⇒ ? = 173 - 128.43 - 30.21
⇒ ? = 173 - 158.64
⇒ ? = 14.36

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

$$\eqalign{ & \left( {{a^4} + \frac{1}{{{a^4}}}} \right){\text{ = 1154}} \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^4} + \frac{1}{{{a^4}}} + 2 = 1156 \cr & \Rightarrow {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = 1156 \cr & \Rightarrow \left( {{a^2} + \frac{1}{{{a^2}}}} \right) = 34 \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 36 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^2} = 36 \cr & \Rightarrow a + \frac{1}{a} = 6 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^3} = {6^3} = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 6 = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 216 - 18 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 198 \cr} $$

Given,
24.96² ÷ (34.11 ÷ 20.05) + 67.96 - 89.11
≈ (25)² ÷ (34 ÷ 20) + 68 - 89
≈ (25)² ÷ $$\frac{{34}}{{20}}$$ + 68 - 89
≈ 625 ÷ 1.7 + 68 - 89
≈ 367.6 + 68 - 89
≈ 367 + 68 - 89
≈ 346

$$\eqalign{ & {\text{Given xepression ,}} \cr & \frac{{\left( {x + y - z} \right)\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)\left( {x + z - y} \right)}} + \frac{{\left( {y + x - z} \right)\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)\left( {x + y - z} \right)}} + \frac{{\left( {z + x - y} \right)\left( {z - x + y} \right)}}{{\left( {y + z + x} \right)\left( {y + z - x} \right)}} \cr & = \frac{{\left( {x + y - z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{\left( {x + y - z} \right) + \left( {y - x + z} \right) + \left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$

$$\eqalign{ & \left( {x + \frac{1}{x}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{2}}^2} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 4 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} \cr & \,\,\,\,\,\,\,\, = 2 - 2 = 0 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = 0}} \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$

1² + 3² + 5² + . . . . . + 17²
= (1² + 2² + 3² + 4² + . . . . . + 17²) - (2² + 4² + . . . . . + 16²)
= (1² + 2² + 3² + 4² + . . . . . + 17²) - 2²(1² + 2² + . . . . . + 8²)
Using given formula
$$\eqalign{ & \left[ {\frac{{p\left( {p + 1} \right)\left( {2p + 1} \right)}}{6}} \right] \cr & = \frac{{17 \times 18 \times 35}}{6} - 4\left[ {\frac{{8 \times 9 \times 17}}{6}} \right] \cr & = 1785 - 4 \times 204 \cr & = 1785 - 816 \cr & = 969 \cr} $$

$$\eqalign{ & \left( {x + \frac{1}{x}} \right) = \sqrt {13} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} - 4 = {\left( {\sqrt {13} } \right)^2} - 4 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow \left( {x - \frac{1}{x}} \right) = 3 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {3^3} = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.x.\frac{1}{x}\left( {x - \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times 3 = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 27 + 9 = 36 \cr} $$

$$\eqalign{ & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\,\,\,\, \cr & \Rightarrow aqr + bpr + cpq = 0....({\text{i}}) \cr & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\,\, \cr & \Rightarrow {\left( {\frac{p}{a} + \frac{q}{b} + \frac{r}{c}} \right)^2} = 1 \cr & \Rightarrow {\text{ }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}\, + 2\left( {\frac{{pq}}{{ab}} + \frac{{pr}}{{ac}} + \frac{{qr}}{{bc}}} \right) = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} + \frac{{2\left( {pqc + prb + qra} \right)}}{{abc}} = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1....\left[ {{\text{using (i)}}} \right] \cr} $$

$$\left( {\frac{1}{{\sqrt 9 - \sqrt 8 }} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }}} \right)$$
$$\eqalign{ & = \sqrt 9 + \sqrt 8 - \sqrt 8 - \sqrt 7 + \sqrt 7 + \sqrt 6 - \sqrt 6 - \sqrt 5 \cr & = \sqrt 9 - \sqrt 5 \cr & = 3 - \sqrt 5 \cr} $$

$$\eqalign{ & {\text{Given experssion ,}} \cr & \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & \left( {Let,\,38 = a,\,34 = b,\,28 = c} \right) \cr & = a + b + c \cr & = 38 + 34 + 28 \cr & = 100 \cr} $$

$$\eqalign{ & {\text{Greatest 4 digit number}} \cr & {\text{ = 9999 }} \cr & \,\,\,\,\,\,\,\,{\text{9| }}\,{\text{9999 | 99}} \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,81 \cr & \overline {189\,\,|\,\,\,1899} \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,1701 \cr & \,\,\,\,\,\,\,\,\,\,\,\overline {|\,\,\,\,\,198} \cr} $$
Greatest four digit number is a perfect square
= 999 - 198
= 9801

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\sqrt {32} + \sqrt {48} }}{{\sqrt 8 + \sqrt {12} }} \cr & = \frac{{\sqrt {4 \times 4 \times 2} + \sqrt {4 \times 4 \times 3} }}{{\sqrt {2 \times 2 \times 2} + \sqrt {2 \times 2 \times 3} }} \cr & = \frac{{4\sqrt 2 + 4\sqrt 3 }}{{2\sqrt 2 + 2\sqrt 3 }} \cr & = \frac{{4\left( {\sqrt 2 + \sqrt 3 } \right)}}{{2\left( {\sqrt 2 + \sqrt 3 } \right)}} \cr & = \frac{4}{2} \cr & = 2 \cr} $$