a: \(\dfrac{16}{15}\cdot\dfrac{-5}{14}\cdot\dfrac{54}{24}\cdot\dfrac{56}{21}\)

\(=\dfrac{16}{24}\cdot\dfrac{-5}{15}\cdot\dfrac{54}{21}\cdot\dfrac{56}{14}\)

\(=\dfrac{2}{3}\cdot\dfrac{-1}{3}\cdot\dfrac{18}{7}\cdot4\)

\(=\dfrac{-2\cdot2\cdot4}{7}=-\dfrac{16}{7}\)

b: \(\dfrac{7}{3}\cdot\dfrac{-5}{2}\cdot\dfrac{15}{21}\cdot\dfrac{4}{-5}\)

\(=\dfrac{7}{3}\cdot\dfrac{5}{7}\cdot\dfrac{5}{2}\cdot\dfrac{4}{5}\)

\(=\dfrac{5}{3}\cdot\dfrac{4}{2}=\dfrac{5}{3}\cdot2=\dfrac{10}{3}\)

c: \(\left(-\dfrac{2}{5}+\dfrac{1}{3}\right)\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)

\(=\dfrac{-2\cdot5+3}{15}\cdot\dfrac{3\cdot7-3\cdot2}{14}\)

\(=\dfrac{-7}{15}\cdot\dfrac{15}{14}=\dfrac{-7}{14}=-\dfrac{1}{2}\)

d: \(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\left(5-\dfrac{1}{4}\right)\)

\(=\dfrac{3-2}{6}\cdot\dfrac{20-1}{4}\)

\(=\dfrac{1}{6}\cdot\dfrac{19}{4}=\dfrac{19}{24}\)

e: \(\left(\dfrac{4}{5}\right)^2=\dfrac{4^2}{5^2}=\dfrac{16}{25}\)

a: \(\left(-\dfrac{3}{7}\right)^2=\dfrac{\left(-3\right)^2}{7^2}=\dfrac{9}{49}\)

b: \(\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)

\(=\dfrac{5}{17}-\dfrac{15}{5}\cdot\dfrac{2}{34}\)

\(=\dfrac{5}{17}-\dfrac{1}{17}\cdot3=\dfrac{2}{17}\)

c: \(\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(=-\dfrac{1}{3}+1+\dfrac{1}{3}\)

=1

d: \(\left(-\dfrac{1}{5}+\dfrac{3}{12}\right)+\dfrac{-3}{4}\)

\(=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}\)

\(=\dfrac{-1}{5}-\dfrac{2}{4}\)

\(=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

a)

\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)                                    

b)

\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)

c)

\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)          

 d)

\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)

e)

 \(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)                                   

 g)

\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)

`@` `\text {Ans}`

`\downarrow`

`a.`

\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-0,4+1\)

`= -0,1 + 1`

`= 0,9`

`b.`

\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)

`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)

`=`\(1+4\cdot\left(-2,25\right)\)

`= 1+ (-9) = -8`

`c.`

\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)

`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)

`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)

`=`\(\dfrac{13}{6}\div2\)

`=`\(\dfrac{13}{12}\)

`d.`

\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)

`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)

`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)

`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)

`e.`

\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)

`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)

`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)

`f.`

\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)

`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)

`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)

`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)

`= 0,8 \div (-2/15)`

`=-6`

`@` `yHGiangg.`

623/12

1/15

5/12

2/7

4/15

1/4

=1/4

a) $\frac{5}{9}:\frac{7}{3} = \frac{5}{9} \times \frac{3}{7} = \frac{{15}}{{63}} = \frac{5}{{21}}$                                           

b) $\frac{2}{7}:\frac{8}{5} = \frac{2}{7} \times \frac{5}{8} = \frac{{10}}{{56}} = \frac{5}{{28}}$

c) $\frac{4}{{15}}:\frac{2}{5} = \frac{4}{{15}} \times \frac{5}{2} = \frac{{20}}{{30}} = \frac{2}{3}$                                         

d) $\frac{7}{9}:\frac{{21}}{9} = \frac{7}{9} \times \frac{9}{{21}} = \frac{{63}}{{189}} = \frac{1}{3}$

a) \(\dfrac{6}{7}+\dfrac{7}{8}=\dfrac{48}{56}+\dfrac{49}{56}=\dfrac{97}{56}\)

b) \(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\)

c) \(\dfrac{2}{3}.\dfrac{4}{9}=\dfrac{8}{27}\)

d) \(\dfrac{1}{5}:\dfrac{2}{7}=\dfrac{1}{5}.\dfrac{7}{2}=\dfrac{7}{10}\)

a: =48/56+49/56

=97/56

b: =12/15-10/15

=2/15

c: =(2*4)/(3*9)=8/27

d: =1/5*7/2=7/10

a: \(=\dfrac{13}{3}+\dfrac{17}{6}=\dfrac{26}{6}+\dfrac{17}{6}=\dfrac{43}{6}\)

b: \(=7-\dfrac{8}{3}=\dfrac{21-8}{3}=\dfrac{13}{3}\)

c: \(=\dfrac{17}{7}\cdot\dfrac{7}{4}=\dfrac{17}{4}\)

d: \(=\dfrac{16}{3}:\dfrac{16}{5}=\dfrac{16}{3}\cdot\dfrac{5}{16}=\dfrac{5}{3}\)