(1,0 điểm) So sánh \(A = \frac{{1 + {{2022}^{2022}}}}{{1 + {{2022}^{2023}}}}\) và \(B = \frac{{1 + {{2022}^{2023}}}}{{1 + {{2022}^{2024}}}}\).

b) Ta thấy \({\widehat M_1} = {\widehat N_1} = 55^\circ \) mà \({\widehat M_1}\) và \({\widehat N_1}\) ở vị trí đồng vị.

Do đó \[MQ\parallel NP\].

Vì \({\widehat M_1}\) và \({\widehat M_2}\) là hai góc kề bù nên \({\widehat M_1} + {\widehat M_2} = 180^\circ \).

Suy ra \({\widehat M_2} = 180^\circ - {\widehat M_1} = 180^\circ - 55^\circ = 125^\circ \).

Ta thấy \({\widehat M_2} = {\widehat Q_1} = 125^\circ \) mà \({\widehat M_2}\) và \({\widehat Q_1}\) ở vị trí so le trong.

Do đó \[MN\parallel PQ\].

Vậy \[MQ\parallel NP & ;\,\,MN\parallel PQ\].

c) Vì \[MQ\parallel NP\] nên \[{\widehat P_1} = {\widehat Q_1} = 125^\circ \] (hai góc đồng vị)

Vì \({\widehat P_1}\) và \({\widehat P_2}\) là hai góc kề bù nên \[{\widehat P_1} + {\widehat P_2} = 180^\circ \].

Suy ra \[{\widehat P_2} = 180^\circ - {\widehat P_1} = 180^\circ - 125^\circ = 55^\circ \].

Vậy \[{\widehat P_2} = 55^\circ \].

a) \(\frac{3}{4} - \frac{2}{3}:x = \frac{{ - 1}}{{20}}\)

\(\frac{2}{3}:x = \frac{3}{4} - \frac{{ - 1}}{{20}}\)

\(\frac{2}{3}:x = \frac{4}{5}\)

\(x = \frac{2}{3}:\frac{4}{5}\)

\(x = \frac{5}{6}\)

Vậy \(x = \frac{5}{6}\).

b) \(\left| {2x - \frac{5}{6}} \right| - \frac{7}{3} = \frac{{ - 11}}{{18}}\)

\(\left| {2x - \frac{5}{6}} \right| = \frac{{ - 11}}{{18}} + \frac{7}{3}\)

\(\left| {2x - \frac{5}{6}} \right| = \frac{{31}}{{18}}\)

TH1: \(2x - \frac{5}{6} = \frac{{31}}{{18}}\)

\(2x = \frac{{31}}{{18}} + \frac{5}{6}\)

\(2x = \frac{{23}}{9}\)

\(x = \frac{{23}}{{18}}\).

TH2: \(2x - \frac{5}{6} = \frac{{ - 31}}{{18}}\)

\(2x = \frac{{ - 31}}{{18}} + \frac{5}{6}\)

\(2x = \frac{{ - 8}}{9}\)

\(x = \frac{{ - 4}}{9}\).

Vậy \(x \in \left\{ {\frac{{23}}{{18}};\,\,\frac{{ - 4}}{9}} \right\}\).

c) \(2{(5 - 4x)^2} - 17 = \frac{{ - 55}}{9}\)

\[2{(5 - 4x)^2} = 17 - \frac{{55}}{9}\]

\[2{(5 - 4x)^2} = \frac{{98}}{9}\]

\[{(5 - 4x)^2} = \frac{{49}}{9}\]

\[{(5 - 4x)^2} = {\left( {\frac{7}{3}} \right)^2}\]

TH1: \[5 - 4x = \frac{7}{3}\]

\[4x = 5 - \frac{7}{3}\]

\[4x = \frac{8}{3}\]

\[x = \frac{2}{3}\].

TH2: \[5 - 4x = \frac{{ - 7}}{3}\]

\[4x = 5 + \frac{7}{3}\]

\[4x = \frac{{22}}{3}\]

\[x = \frac{{11}}{6}\].

Vậy \[x \in \left\{ {\frac{2}{3};\,\,\frac{{11}}{6}} \right\}\].