a: \(=\dfrac{6}{7}\cdot\dfrac{-3}{5}=\dfrac{-18}{35}\)

b: \(=\dfrac{2}{5}\cdot\dfrac{-15}{8}=\dfrac{-30}{40}=-\dfrac{3}{4}\)

c: \(=\dfrac{2}{4}\cdot\dfrac{7}{3}=\dfrac{1}{2}\cdot\dfrac{7}{3}=\dfrac{7}{6}\)

d: \(=\dfrac{8}{3}\cdot\dfrac{16}{4}=\dfrac{128}{12}=\dfrac{32}{3}\)

a)-18/35

b)-3/4

c)7/6

d)32/3

e)5/21

sai thì xin lỗi nhé bn

a) \( - 2{x^2} + 6{x^2} = ( - 2 + 6).{x^2} = 4{x^2}\);                  

b) \(4{x^3} - 8{x^3} = (4 - 8).{x^3} =  - 4{x^3}\);

c) \(3{x^4}( - 6{x^2}) = 3.( - 6).{x^4}.{x^2} =  - 18{x^{4 + 2}} =  - 18{x^6}\);                 

d) \(( - 24{x^6}):( - 4{x^3}) = ( - 24: - 4).({x^6}:{x^3}) = 6{x^{6 - 3}} = 6{x^3}\). 

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

` 6/7 + 7/8 = 48/56 + 49/56 = 97/56`

` 4/5 - 2/3 = 12/15 - 10/15 = 2/15`

` 2/3 xx 4/9 = 8/27`

` 1/5 : 2/7 = 1/5 . 7/2 = 7/10`

Tham khảo:

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: \(\dfrac{3}{8}+\dfrac{7}{8}=\dfrac{3+7}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

b: \(\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{7-4}{9}=\dfrac{3}{9}=\dfrac{1}{3}\)

c: \(\dfrac{5}{6}+\dfrac{1}{8}=\dfrac{20}{24}+\dfrac{3}{24}=\dfrac{20+3}{24}=\dfrac{23}{24}\)

d: \(\dfrac{9}{15}-\dfrac{2}{5}=\dfrac{3}{5}-\dfrac{2}{5}=\dfrac{3-2}{5}=\dfrac{1}{5}\)

e: \(\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{9}{5}=\dfrac{2+3+9}{5}=\dfrac{14}{5}\)

g: \(\dfrac{8}{10}-\dfrac{1}{10}-\dfrac{3}{10}=\dfrac{8-1-3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

h: \(\dfrac{23}{7}-\dfrac{4}{7}+\dfrac{2}{7}=\dfrac{23-4+2}{7}=\dfrac{21}{7}=3\)

a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)

\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)

b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)

a) $\frac{3}{2} \times \frac{5}{8} + \frac{7}{4} = \frac{{15}}{{16}} + \frac{7}{4} = \frac{{15}}{{16}} + \frac{{28}}{{16}} = \frac{{43}}{{16}}$

b) $\frac{8}{5}:\left( {\frac{4}{3} - \frac{5}{6}} \right) = \frac{8}{5}:\left( {\frac{8}{6} - \frac{5}{6}} \right) = \frac{8}{5}:\frac{1}{2} = \frac{8}{5} \times 2 = \frac{{16}}{5}$

c) $\frac{3}{4} \times \frac{1}{5} - \frac{1}{{10}} = \frac{3}{{20}} - \frac{1}{{10}} = \frac{3}{{20}} - \frac{2}{{20}} = \frac{1}{{20}}$

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)

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