Simplification

$${\text{Given expression,}}$$
$$\frac{{28 + 14 + 7 + 4 + 2 + 1}}{{28}}$$
$$ = \frac{{56}}{{28}}$$
$$ = 2$$

$$\frac{1}{x} = 4 - \left( {\frac{1}{3} + \frac{1}{2}} \right)$$
$$\,\,\,\,\,\, = 4 - {\frac{{2 + 3}}{6}} $$
$$\,\,\,\,\,\, = 4 - \frac{5}{6}$$
$$\,\,\,\,\,\, = \frac{{24 - 5}}{6}$$
$$\,\,\,\,\,\, = \frac{{19}}{6}$$
$$ ⇒ x = \frac{6}{{19}}$$

$${\text{we know that,}}$$
$$\frac{{\sqrt {80} - \sqrt {112} }}{{\sqrt {45} - \sqrt {63} }}$$
$$ ⇒ \frac{{4\sqrt 5 - 4\sqrt 7 }}{{3\sqrt 5 - 3\sqrt 7 }}$$
$$ ⇒ \frac{{4\left( {\sqrt 5 - \sqrt 7 } \right)}}{{3\left( {\sqrt 5 - \sqrt 7 } \right)}}$$
$$ ⇒ \frac{4}{3}$$
$$ ⇒ 1\frac{1}{3}$$

$${\text{we know that}} \cr &×+ y = 22\,......(i)$$
$${x^2} + {y^2} = 404\,......(ii){\text{ }}$$
$$\therefore {\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$$
$$ ⇒ {\left( {22} \right)^2} = 404 + 2xy$$
$$ ⇒ 484 = 404 + 2xy$$
$$ ⇒ 2xy = 80$$
$$ ⇒ xy = 40$$

$${\text{we know that,}}$$
$${\text{8}}\frac{1}{2} - \left[ {3\frac{1}{4} + \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]$$
$$ ⇒ \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]$$
$$ ⇒ \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{{9 - 2 - 1}}{6}} \right)} \right\}} \right]$$
$$ ⇒ \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2} \times 1} \right\}} \right]$$
$$ ⇒ \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{{5 - 2}}{4}} \right\}} \right]$$
$$ ⇒ \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \frac{3}{4}} \right]$$
$$ ⇒ \frac{{17}}{2} - \frac{{16}}{4}$$
$$ ⇒ \frac{{34 - 16}}{4}$$
$$ ⇒ \frac{9}{2}$$
$$ ⇒ 4\frac{1}{2}$$

$$1 + \frac{1}{{1 + \frac{2}{{2 + \frac{3}{{1 + \frac{4}{5}}}}}}}$$
$$ = 1 + \frac{1}{{1 + \frac{2}{{2 + \frac{3}{{\frac{9}{5}}}}}}}$$
$$ = 1 + \frac{1}{{1 + \frac{2}{{2 + \frac{5}{3}}}}}$$
$$ = 1 + \frac{1}{{1 + \frac{2}{{\frac{{11}}{3}}}}}$$
$$ = 1 + \frac{1}{{1 + \frac{6}{{11}}}}$$
$$ = 1 + \frac{1}{{\frac{{17}}{{11}}}}$$
$$ = 1 + \frac{{11}}{{17}}$$
$$ = 1\frac{{11}}{{17}}$$

$$1 + \frac{2}{{1 + \frac{3}{{1 + \frac{4}{5}}}}}$$
$$ = 1 + \frac{2}{{1 + \frac{3}{{\frac{9}{5}}}}}$$
$$ = 1 + \frac{2}{{1 + \frac{5}{3}}}$$
$$ = 1 + \frac{2}{{\frac{8}{3}}}$$
$$ = 1 + \frac{3}{4}$$
$$ = \frac{7}{4}$$

$${\text{According to question}}$$
$$\frac{{9|3 - 5| - 5|4| \div 10}}{{ - 3\left( 5 \right) - 2 \times 4 \div 2}}$$
$$ ⇒ \frac{{9 \times 2 - 20 \div 10}}{{ - 15 - 2 \times 2}}$$
$$ ⇒ \frac{{18 - 2}}{{ - 15 - 4}}$$
$$ ⇒ - \frac{{16}}{{19}}$$

Given,
5 - [4 - {3 - (3 - 3 - 6)}]
= 5 - [4 - {3 - (-6)}]
= 5 - [4 - {3 +6}]
= 5 - [4 - {9}]
= 5 - [4 - 9]
= 5 - [-5]
= 5 + 5
= 10

$$\frac{{ - {{\left( {4 - 6} \right)}^2} - 3\left( { - 2} \right) + \left| { - 6} \right|}}{{18 - 9 \div 3 \times 5}}$$
$$ = \frac{{ - {{\left( { - 2} \right)}^2} - \left( { - 6} \right) + 6}}{{18 - 3 \times 5}}$$
$$ = \frac{{ - 4 + 6 + 6}}{{18 - 15}}$$
$$ = \frac{8}{3}$$