Hướng dẫn giải

a) \(\frac{{31}}{{23}} - \left( {\frac{7}{{32}} + \frac{8}{{23}}} \right)\)

\( = \frac{{31}}{{23}} - \frac{7}{{32}} - \frac{8}{{23}}\)

\( = \left( {\frac{{31}}{{23}} - \frac{8}{{23}}} \right) - \frac{7}{{32}}\)

\( = \frac{{23}}{{23}} - \frac{7}{{32}}\)

\( = 1 - \frac{7}{{32}}\)

\( = \frac{{25}}{{32}}.\)

b) \(\left( {1 - \frac{2}{3} - \frac{1}{4}} \right):{\left( {\frac{4}{5} - \frac{3}{4}} \right)^2}\)

\( = \left( {\frac{{12}}{{12}} - \frac{8}{{12}} - \frac{3}{{12}}} \right):{\left( {\frac{{16}}{{20}} - \frac{{15}}{{20}}} \right)^2}\)

\( = \frac{1}{{12}}:{\left( {\frac{1}{{20}}} \right)^2}\)

\( = \frac{1}{{12}}:\frac{1}{{400}}\)

\( = \frac{1}{{12}}.400\)

\( = \frac{{400}}{{12}}\)

\( = \frac{{100}}{3}.\)

c) \({\left( { - \frac{1}{2}} \right)^2}.\frac{2}{3} + \sqrt {81} .\left( {\frac{{ - 2}}{3}} \right) + 1\frac{1}{2}.\sqrt {\frac{4}{9}} \)

\( = \frac{1}{4}.\frac{2}{3} - \sqrt {{9^2}} .\frac{2}{3} + \frac{3}{2}.\sqrt {{{\left( {\frac{2}{3}} \right)}^2}} \)

\( = \frac{1}{4}.\frac{2}{3} - 9.\frac{2}{3} + \frac{3}{2}.\frac{2}{3}\)

\( = \left( {\frac{1}{4} - 9 + \frac{3}{2}} \right).\frac{2}{3}\)

\( = \left( {\frac{1}{4} - \frac{{36}}{4} + \frac{6}{4}} \right).\frac{2}{3}\)

\( = - \frac{{29}}{4}.\frac{2}{3}\)

\( = - \frac{{29}}{6}\).

Số √ 5 là