(2,0 điểm)

1. Thực hiện phép tính (tính nhanh nếu có thể)

(a) \(\frac{{ - 2}}{3}\,\,.\,\,2 + \frac{4}{5}:3\)

(b) \({\left( {\frac{{ - 3}}{4}} \right)^5}:{\left( {\frac{{ - 3}}{4}} \right)^4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{8^5}\,\,.\,\,{9^{10}}}}\).

2. Tìm \(x\), biết:

(a) \({3^3} - 0,5x = 26,75\)

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

1.

a) \(\frac{{ - 2}}{3}\,\,.\,\,2 + \frac{4}{5}:3 = \frac{{ - 4}}{3} + \frac{4}{5}\,\,.\,\,\frac{1}{3} = \frac{{ - 4}}{3} + \frac{4}{{15}} = \frac{{ - 16}}{{15}}\);

b) \({\left( {\frac{{ - 3}}{4}} \right)^5}:{\left( {\frac{{ - 3}}{4}} \right)^4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{8^5}\,\,.\,\,{9^{10}}}} = \frac{{ - 3}}{4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{{\left( {{2^3}} \right)}^5}\,\,.\,\,{{\left( {{3^2}} \right)}^{10}}}}\)

\( = \frac{{ - 3}}{4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{2^{15}}\,\,.\,\,{3^{20}}}} = \frac{{ - 3}}{4} + {3^2} = \frac{{ - 3}}{4} + 9 = \frac{{33}}{4}\).

2.

a) \({3^3} - 0,5x = 26,75\)

\(27 - 0,5x = 26,75\)

\(0,5x = 27 - 26,75\)

\(0,5x = 0,25\)

\(x = 0,5\)

Vậy \(x = 0,5\).

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

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

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

\[\left| {x - \frac{1}{3}} \right| \cdot 2 = 2\]

\[\left| {x - \frac{1}{3}} \right| = 2:2\]

\[\left| {x - \frac{1}{3}} \right| = 1\]

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 \[x \in \left\{ {\frac{4}{3};\,\,\frac{{ - 2}}{3}} \right\}\].