Cho hai đường thẳng song song a và b như hình vẽ.

Hai góc ở vị trí so le trong là:

1.

a) \({\left( { - \frac{1}{3}} \right)^2} + 2\,\,.\,\,\frac{3}{4} - \frac{1}{2} = \frac{1}{{{3^2}}} + \frac{3}{2} - \frac{1}{2}\)\( = \frac{1}{9} + \frac{2}{2} = \frac{1}{9} + 1 = \frac{{10}}{{9\;}}\);

b) \({\left( {\frac{3}{4}} \right)^2} - \frac{{22}}{8} + \left[ {\left( { - \frac{5}{2}} \right) - \left( {5 - \frac{2}{4}} \right)} \right]\)\( = \frac{{{3^2}}}{{{4^2}}} - \frac{{11}}{4} + \left[ {\frac{{ - 5}}{2} - \left( {5 - \frac{1}{2}} \right)} \right]\)

\( = \frac{9}{{16}} - \frac{{11}}{4} + \left[ {\frac{{ - 5}}{2} - \frac{9}{2}} \right] = \frac{9}{{16}} - \frac{{11}}{4} - 7 = \frac{{ - 35}}{{16}} - 7\)\( = \frac{{ - 147}}{{16}}\).

2.

a) \(1,5.\frac{2}{3}:x = \frac{7}{9}:0,5\)      

\(\frac{{1,5.2}}{3}:x = \frac{7}{9}:\frac{1}{2}\)

\(1:x = \frac{{14}}{9}\)

\(x = 1:\frac{{14}}{9}\)

\(x = \frac{9}{{14}}\).

Vậy \(x = \frac{9}{{14}}\).

b) \(\left| {x - \frac{1}{3}} \right| \cdot 2 - 2\frac{1}{9} = - {\left( { - \frac{1}{3}} \right)^2}\)

\(\left| {x - \frac{1}{3}} \right| \cdot 2 - \frac{{19}}{9} = - \frac{1}{9}\)

\(\left| {x - \frac{1}{3}} \right| \cdot 2 = \left( { - \frac{1}{9}} \right) + \frac{{19}}{9}\)

\(\left| {x - \frac{1}{3}} \right| \cdot 2 = 2\)

\(\left| {x - \frac{1}{3}} \right| = 1\)

TH1: \(x - \frac{1}{3} = 1\)

\(x = 1 + \frac{1}{3}\)

\(x = \frac{4}{3}\).

TH2: \(x - \frac{1}{3} = - 1\)

\(x = - 1 + \frac{1}{3}\)

\(x = - \frac{2}{3}\).

Vậy \(x \in \left\{ {\frac{4}{3};\,\,\frac{{ - 2}}{3}} \right\}\).

Ta có \(B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{2}{{2020}} + \frac{1}{{2021}}\)

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

\( = 1 + \frac{{2020 + 2}}{2} + \frac{{2019 + 3}}{3} + ... + \frac{{2 + 2020}}{{2020}} + \frac{{1 + 2021}}{{2021}}\)

\( = \frac{{2022}}{{2022}} + \frac{{2022}}{2} + \frac{{2022}}{3} + ... + \frac{{2022}}{{2022}} + \frac{{2022}}{{2021}}\)

\( = 2022\left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2020}} + \frac{1}{{2021}} + \frac{1}{{2022}}} \right) = 2022A\).

Vậy \(\frac{A}{B} = \frac{A}{{2022A}} = \frac{1}{{2022}}\).