(1,5 điểm) Cho hình vẽ, biết \[AB\parallel xy\], \(\widehat {BAI} = 45^\circ \), \(\widehat {AIF} = 105^\circ \).

(a) Vẽ lại hình và viết giả thiết, kết luận của bài toán.

(b) Tính số đo \(\widehat {FIx}\) và \(\widehat {FIy}\).

(c) Chứng minh \[AB\parallel EF\].

a) \[\frac{4}{3}:\frac{{ - 5}}{6} + \frac{{11}}{2} = \frac{{ - 8}}{5} + \frac{{11}}{2} = \frac{{39}}{{10}}\];

b) \(\frac{3}{4}\,\,.\,\,\frac{{ - 4}}{7} + \frac{{ - 4}}{7}\,\,.\,\,\frac{5}{4} = \frac{{ - 4}}{7}\,\,.\,\,\left( {\frac{5}{4} + \frac{3}{4}} \right) = \frac{{ - 4}}{7}\,\,.\,\,2 = \frac{{ - 8}}{7}\);

c) \[\sqrt {0,81} \,\,.\,\,{\left( {\frac{{ - 1}}{3}} \right)^2} - \sqrt {\frac{1}{{196}}} \,\,.\,\,{( - 7)^3} = 0,9\,\,.\,\,\frac{1}{9} + \frac{1}{{14}}\,\,.\,\,{7^3}\]

\[ = 0,9\,\,.\,\,\frac{1}{9} + \frac{1}{{14}}\,\,.\,\,{7^3} = \frac{1}{{10}} + \frac{{{7^3}}}{{7\,\,.\,\,2}} = \frac{1}{{10}} + \frac{{49}}{2} = \frac{{123}}{5}\].

a) \(\frac{4}{5}x + \frac{5}{6} = \frac{{41}}{{30}}\)

\(\frac{4}{5}x = \frac{{41}}{{30}} - \frac{5}{6}\)

\(\frac{4}{5}x = \frac{8}{{15}}\)

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

Vậy \(x = \frac{2}{3}\).

b) \(\left| {2x + \frac{4}{3}} \right| = {\left( {0,45} \right)^{2023}}:{\left( {0,45} \right)^{2022}}\)

\(\left| {2x + \frac{4}{3}} \right| = {\left( {0,45} \right)^1}\)

\(\left| {2x + \frac{4}{3}} \right| = \frac{3}{{20}}\)

TH1: \(2x + \frac{4}{3} = \frac{3}{{20}}\)

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

\(2x = \frac{{ - 71}}{{60}}\)

\(x = \frac{{ - 71}}{{120}}\).

Vậy \(x = \frac{{ - 71}}{{120}}\).

TH2: \(2x + \frac{4}{3} = \frac{{ - 3}}{{20}}\)

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

\(2x = \frac{{ - 89}}{{60}}\)

\(x = \frac{{ - 89}}{{120}}\).

Vậy \(x \in \left\{ {\frac{{ - 71}}{{120}};\,\,\frac{{ - 89}}{{120}}} \right\}\).

c) \(5 + \sqrt x = \frac{5}{2} + \frac{{27}}{8}\)

\(5 + \sqrt x = \frac{{47}}{8}\)

\(\sqrt x = \frac{{47}}{8} - 5\)

\(\sqrt x = \frac{7}{8}\)

\(x = \frac{{49}}{{64}}\).

Vậy \(x = \frac{{49}}{{64}}\).