Decimal Fraction

Given expression :
$$ = \frac{{{{\left( {0.943} \right)}^2} - \left( {0.943 \times 0.057} \right) + {{\left( {0.057} \right)}^2}}}{{{{\left( {0.943} \right)}^3} + {{\left( {0.057} \right)}^3}}}$$
$$ = \frac{{{a^2} - ab + {b^2}}}{{{a^3} + {b^3}}}$$
$$ = \frac{1}{{a + b}}$$
$$ = \frac{1}{{0.943 + 0.057}}$$
$$ = 1$$

Let, $$\frac{{{x^2} + {{\left( {18} \right)}^2}}}{{125}} = 3.56$$
Then,
$${x^2} + 324 = 125 \times 3.56 = 445$$
$$ ⇒ {x^2} = 121$$
$$ ⇒ x = 11$$

Let 534.596 + 61.472 - 496.708 = x + 27.271
Then,×= (534.596 + 61.472) - (496.708 + 27.271)
= 596.068 - 523.979
= 72.089

$${\left( {\frac{{18}}{4}} \right)^2} \times \left( {\frac{{455}}{{19}}} \right) \div \left( {\frac{{61}}{{799}}} \right)$$
$$ = \frac{{324}}{{16}} \times \frac{{455}}{{19}} \times \frac{{799}}{{61}}$$
$$ = 6350$$

Let the missing number be x
$$\frac{{294 \div 14 \times 5 + 11}}{x} = {8^2} \div 5 + 1.7$$
$$ ⇒ \frac{{\frac{{294}}{{14}} \times 5 + 11}}{x} = \frac{{64}}{5} + 1.7$$
$$ ⇒ \frac{{21 \times 5 + 11}}{x} = 12.8 + 1.7$$
$$ ⇒ \frac{{105 + 11}}{x} = 12.8 + 1.7$$
$$ ⇒ \frac{{116}}{x} = 14.5$$
$$ ⇒ x = \frac{{116}}{{14.5}}$$
$$ ⇒ x = \frac{{116 \times 10}}{{145}}$$
$$ ⇒ x = 8$$
Hence, the number is 8

$$\frac{3}{4} = 0.75,\  $$
$$\frac{5}{6} = 0.833,\  $$
$$\frac{1}{2} = 0.5,$$
$$\frac{2}{3} = 0.66,$$
$$\  \frac{4}{5} = 0.8,$$
$$\frac{9}{{10}} = 0.9$$
Clearly, 0.8 lies between 0.75 and 0.833
$$\therefore \frac{4}{5}{\text{lies between}}\frac{3}{4}{\text{and}}\frac{5}{6}$$

$$0.125125... = 0.\overline {125} = \frac{{125}}{{999}}$$

$$617.00$$
$$\,\,\,\,\,\,\,6.017$$
$$\,\,\,\,\,\,\,0.617$$
$$ + \,\,\,6.0017$$
$$ - - - - - - $$
$$\,\,\,629.6357$$
$$ - - - - - - $$

$$\frac{{489.1375 \times 0.0483 \times 1.956}}{{0.0873 \times 92.581 \times 99.749}} \approx \frac{{489 \times 0.05 \times 2}}{{0.09 \times 93 \times 100}}$$
$$ = \frac{{489}}{{9 \times 93 \times 10}}$$
$$ = \frac{{163}}{{279}} \times \frac{1}{{10}}$$
$$ = \frac{{0.58}}{{10}}$$
$$ = 0.058 \approx 0.06$$

2×5 = 10.
Sum of decimal places = 4
∴ 0.002×0.5 = 0.001