1.

a) \(\left( {2\frac{1}{5} - \frac{2}{5}} \right) \cdot \left[ {1\frac{1}{3} \cdot \left( { - 5} \right) - \frac{4}{3} + \frac{7}{9}:\frac{1}{6}} \right]\)

\( = \left( {\frac{{11}}{5} - \frac{2}{5}} \right) \cdot \left[ {\frac{4}{3} \cdot \left( { - 5} \right) - \frac{4}{3} + \frac{7}{9} \cdot 6} \right]\)

\( = \frac{9}{5} \cdot \left( {\frac{{ - 20}}{3} - \frac{4}{3} + \frac{{14}}{3}} \right)\)\( = \frac{9}{5} \cdot \left( {\frac{{ - 10}}{3}} \right)\)\( = - 6\).

b) \(\sqrt {{{\left( { - \frac{5}{2}} \right)}^2}} \cdot \frac{4}{5} + \sqrt {1\frac{9}{{16}}} :{\left( {2\frac{1}{2}} \right)^2} - \left| {\frac{{ - 3}}{5}} \right| \cdot 2\)

\( = \frac{5}{2} \cdot \frac{4}{5} + \sqrt {\frac{{25}}{{16}}} :{\left( {\frac{5}{2}} \right)^2} - \frac{6}{5}\)\( = 2 + \sqrt {{{\left( {\frac{5}{4}} \right)}^2}} :\frac{{25}}{4} - \frac{6}{5}\)

\( = 2 + \frac{5}{4} \cdot \frac{4}{{25}} - \frac{6}{5}\)\( = 2 + \frac{1}{5} - \frac{6}{5}\)

\( = 2 + \left( {\frac{1}{5} - \frac{6}{5}} \right)\)\( = 2 + \left( { - 1} \right) = 1\).

2.

a) \[\frac{2}{3}x:\frac{1}{3} = 5\frac{1}{2}\]

\[\frac{2}{3}x\,\,.\,\,3 = \frac{{11}}{2}\]

\[2x = \frac{{11}}{2}\]

\[x = \frac{{11}}{2}:2\]

\[x = \frac{{11}}{4}\].

Vậy \[x = \frac{{11}}{4}\].

b) \(2 \cdot \left( {\frac{1}{5} - x} \right) + \left( {\frac{1}{5} - x} \right) \cdot 3x - 1\frac{1}{5} = - \frac{6}{5}\)

\(\left( {\frac{1}{5} - x} \right) \cdot \left( {2 + 3x} \right) = - \frac{6}{5} + 1\frac{1}{5}\)

\(\left( {\frac{1}{5} - x} \right) \cdot \left( {2 + 3x} \right) = \left( { - \frac{6}{5}} \right) + \frac{6}{5}\)

\(\left( {\frac{1}{5} - x} \right) \cdot \left( {2 + 3x} \right) = 0\)

TH1: \(\frac{1}{5} - x = 0\)

Suy ra \(x = \frac{1}{5}\)

TH2: \(2 + 3x = 0\)

\(3x = - 2\)

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

Vậy \(x \in \left\{ {\frac{1}{5};\,\, - \frac{2}{3}} \right\}\).

Giá trị tuyệt đối của - 4,7 là