Question

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Answer 1

a,

$\frac{1}{1 - \frac{1}{1 - \frac{1}{2}}} + \frac{1}{1 + \frac{1}{1 + \frac{1}{2}}} = \frac{1}{1 - \frac{1}{\frac{1}{2}}} + \frac{1}{1 + \frac{1}{\frac{3}{2}}} = \frac{1}{1 - 2} + \frac{1}{1 + \frac{2}{3}} = \frac{1}{- 1} + \frac{1}{\frac{5}{3}} = - 1 + \frac{3}{5} = \frac{- 2}{5}$

b,

$\frac{\frac{1}{3} - \frac{1}{4} + \frac{1}{5}}{\frac{7}{6} - \frac{7}{8} + \frac{7}{10}} = \frac{\frac{1}{3} - \frac{1}{4} + \frac{1}{5}}{\frac{7}{2} \times (\frac{1}{3} - \frac{1}{4} + \frac{1}{5})} = \frac{1}{\frac{7}{2}} = \frac{2}{7}$

c,

$\frac{\frac{19}{17} - \frac{100}{49} + \frac{129}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{\frac{17 + 2}{17} - \frac{49 \times 2 + 2}{49} + \frac{131 - 2}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{1 + \frac{2}{17} - 2 - \frac{2}{49} + 1 - \frac{2}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{(1 - 2 + 1) + \frac{2}{17} - \frac{2}{49} - \frac{2}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{2 \times (\frac{1}{17} - \frac{1}{49} - \frac{1}{131})}{3 \times (\frac{1}{17} - \frac{1}{49} - \frac{1}{131})} = \frac{2}{3}$

d,

$\frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{\frac{2020}{1} + \frac{2019}{2} + \frac{2018}{3} + . . . + \frac{1}{2020}} = \frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{(1 + \frac{2019}{2}) + (1 + \frac{2018}{3}) + . . . + (1 + \frac{1}{2020}) + 1} = \frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{\frac{2021}{2} + \frac{2021}{3} + . . . + \frac{2021}{2020} + \frac{2021}{2021}} = \frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{2021 \times (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021})} = \frac{1}{2021}$

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Answer 2

a,

$\frac{1}{1 - \frac{1}{1 - \frac{1}{2}}} + \frac{1}{1 + \frac{1}{1 + \frac{1}{2}}} = \frac{1}{1 - \frac{1}{\frac{1}{2}}} + \frac{1}{1 + \frac{1}{\frac{3}{2}}} = \frac{1}{1 - 2} + \frac{1}{1 + \frac{2}{3}} = \frac{1}{- 1} + \frac{1}{\frac{5}{3}} = - 1 + \frac{3}{5} = \frac{- 2}{5}$

b,

$\frac{\frac{1}{3} - \frac{1}{4} + \frac{1}{5}}{\frac{7}{6} - \frac{7}{8} + \frac{7}{10}} = \frac{\frac{1}{3} - \frac{1}{4} + \frac{1}{5}}{\frac{7}{2} \times (\frac{1}{3} - \frac{1}{4} + \frac{1}{5})} = \frac{1}{\frac{7}{2}} = \frac{2}{7}$

c,

$\frac{\frac{19}{17} - \frac{100}{49} + \frac{129}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{\frac{17 + 2}{17} - \frac{49 \times 2 + 2}{49} + \frac{131 - 2}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{1 + \frac{2}{17} - 2 - \frac{2}{49} + 1 - \frac{2}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{(1 - 2 + 1) + \frac{2}{17} - \frac{2}{49} - \frac{2}{131}}{\frac{3}{17} - \frac{3}{49} - \frac{3}{131}} = \frac{2 \times (\frac{1}{17} - \frac{1}{49} - \frac{1}{131})}{3 \times (\frac{1}{17} - \frac{1}{49} - \frac{1}{131})} = \frac{2}{3}$

d,

$\frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{\frac{2020}{1} + \frac{2019}{2} + \frac{2018}{3} + . . . + \frac{1}{2020}} = \frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{(1 + \frac{2019}{2}) + (1 + \frac{2018}{3}) + . . . + (1 + \frac{1}{2020}) + 1} = \frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{\frac{2021}{2} + \frac{2021}{3} + . . . + \frac{2021}{2020} + \frac{2021}{2021}} = \frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021}}{2021 \times (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2021})} = \frac{1}{2021}$

Answer 3

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