Tính giá trị của biểu thức:

\[A = \left( {1 + \frac{1}{{1.3}}} \right).\left( {1 + \frac{1}{{2.4}}} \right).\left( {1 + \frac{1}{{3.5}}} \right).\,\,...\,\,.\left( {1 + \frac{1}{{2022.2024}}} \right)\,\].

a) \(1,2x + \frac{1}{2} = 0,6\)

\(1,2x + 0,5 = 0,6\)

\(1,2x = 0,1\)

\(x = \frac{1}{{12}}\).

Vậy \(x = \frac{1}{{12}}\).

b) \(\left( {{x^2} + 1} \right)\left( {3x - 4} \right) = 0\)

Trường hợp 1:

\({x^2} + 1 = 0\)

Vì \({x^2} \ge 0\) nên \({x^2} + 1 > 0\).

Do đó \({x^2} + 1 \ne 0\).

Vậy \(x = \frac{4}{3}\).

Trường hợp 2:

\(3x - 4 = 0\)

\(3x = 4\)

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

a) \(\frac{4}{5} + \frac{3}{5}.\left( { - \frac{2}{{15}}} \right)\)

\( = \frac{4}{5} + \frac{{ - 2}}{{25}}\)

\( = \frac{{20}}{{25}} + \frac{{ - 2}}{{25}}\)

\( = \frac{{18}}{{25}}\);

b) \(\frac{{ - 5}}{{17}}.\frac{{31}}{{33}} - \frac{5}{{17}}.\frac{2}{{33}} + 1\frac{5}{{17}}\)

\( = \frac{{ - 5}}{{17}}.\left( {\frac{{31}}{{33}} + \frac{2}{{33}}} \right) + \frac{{22}}{{17}}\)

\( = \frac{{ - 5}}{{17}}.1 + \frac{{22}}{{17}}\)

\( = \frac{{17}}{{17}} = 1\).

c) \(1\frac{{13}}{{15}}.0,75 - \left( {\frac{{11}}{{20}} + 25\% } \right):\frac{5}{7}\)

\( = \frac{{28}}{{15}}.\frac{3}{4} - \left( {\frac{{11}}{{20}} + \frac{1}{4}} \right).\frac{7}{5}\)

\( = \frac{7}{5} - \left( {\frac{{11}}{{20}} + \frac{5}{{20}}} \right).\frac{7}{5}\)

\( = \frac{7}{5} - \frac{{16}}{{20}}.\frac{7}{5}\)

\( = \frac{7}{5}.\left( {1 - \frac{{16}}{{20}}} \right)\)

\( = \frac{7}{5}.\frac{4}{{20}} = \frac{7}{{25}}\).