a)P=\(\dfrac{-2}{7}+\dfrac{14}{29}+\dfrac{12}{33}+\dfrac{15}{29}+\dfrac{21}{33}+\dfrac{2}{7}\)

      =\(\left(\dfrac{-2}{7}+\dfrac{2}{7}\right)+\left(\dfrac{14}{29}+\dfrac{15}{29}\right)+\left(\dfrac{12}{33}+\dfrac{21}{33}\right)\)

      =0+1+1=2

b)\(\dfrac{2}{7}.\dfrac{5}{19}+\dfrac{2}{7}.\dfrac{14}{19}+\dfrac{21}{19}-\dfrac{2}{7}.\dfrac{1}{5}\)

=\(\dfrac{2}{7}.\left(\dfrac{5}{19}+\dfrac{14}{19}-\dfrac{1}{5}\right)+\dfrac{21}{19}\)

=\(\dfrac{2}{7}.\dfrac{4}{5}+\dfrac{21}{19}=\dfrac{887}{665}\)

\(\dfrac{15}{34}-\dfrac{7}{21}+\dfrac{19}{34}-\dfrac{15}{17}-\dfrac{2}{3}\)

\(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)-\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{15}{17}\)

\(=1-\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{15}{17}\)

\(=1-1-\dfrac{15}{17}\)

\(=-\dfrac{15}{17}\)

Câu 1:

\(=\dfrac{15}{34}+\dfrac{19}{34}-1-\dfrac{15}{17}+\dfrac{1}{3}+\dfrac{3}{5}\)

\(=-\dfrac{15}{17}+\dfrac{14}{15}=\dfrac{13}{255}\)

Câu 2: 

\(=\dfrac{5^4\cdot5^4\cdot2^8}{4^4\cdot6^4\cdot3^2\cdot5}=\dfrac{5^7}{6^4\cdot3^2}\)

a, \(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-\dfrac{20}{15}+\dfrac{3}{7}\)

= \(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{3}{7}\right)-\dfrac{20}{15}\)

= 1 + \(\dfrac{16}{21}-\dfrac{20}{15}\)

= \(\dfrac{3}{7}\)

b, \(\dfrac{27}{25}+\dfrac{4}{21}-\dfrac{2}{25}+\dfrac{17}{21}-\dfrac{1}{2}\)

= \(\left[\dfrac{27}{25}+\left(-\dfrac{2}{25}\right)\right]+\left(\dfrac{4}{21}-\dfrac{7}{21}\right)-\dfrac{1}{2}\)

= 1 + \(\left(\dfrac{-1}{7}\right)-\dfrac{1}{2}\) = \(\dfrac{5}{14}\)

D

a) \(\dfrac{3}{5}+\dfrac{7}{25}=\dfrac{15}{25}+\dfrac{7}{25}=\dfrac{15+7}{25}=\dfrac{22}{25}\)

b) \(\dfrac{8}{11}-\dfrac{19}{33}=\dfrac{24}{33}-\dfrac{19}{33}=\dfrac{24-19}{33}=\dfrac{5}{33}\)

c) \(\dfrac{16}{21}\times\dfrac{3}{5}=\dfrac{16\times3}{21\times5}=\dfrac{48}{105}=\dfrac{16}{35}\)

d) \(\dfrac{14}{41}\div\dfrac{7}{9}=\dfrac{14}{41}\times\dfrac{9}{7}=\dfrac{14\times9}{41\times7}=\dfrac{126}{287}=\dfrac{18}{41}\)

e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)

=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)

=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)

=\(-\dfrac{516}{7}\)

a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)

=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)

=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)

=\(\dfrac{119}{240}\)

b)\(\dfrac{3}{2}-\dfrac{5}{6}:\left(\dfrac{1}{2}\right)^2+\sqrt{4}\)

=\(\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+2\)

=\(\dfrac{3}{2}-\dfrac{10}{3}+2\)

=\(-\dfrac{11}{6}\) +2

=\(\dfrac{1}{6}\)

a: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{2}{17}\)

b: \(=-8\cdot\dfrac{1}{4}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-2:\dfrac{27-14}{12}=\dfrac{-2\cdot12}{13}=-\dfrac{24}{13}\)

c: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)\)

=-520/7

a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A =  - 2 + (-1)\\A =  - 3\end{array}\)

b, 

Cách 1:

\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)

Cách 2:

\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B =  \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)

\(A=\dfrac{15}{34}+\dfrac{27}{21}+\dfrac{9}{34}-1\dfrac{15}{17}+\dfrac{2}{3}\)

\(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1\dfrac{15}{17}\right)+\left(\dfrac{27}{21}+\dfrac{2}{3}\right)\)

\(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-\dfrac{32}{17}\right)+\left(\dfrac{27}{21}+\dfrac{2}{3}\right)\)

\(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-\dfrac{64}{34}\right)+\left(\dfrac{27}{21}+\dfrac{14}{21}\right)\)

\(A=\dfrac{-20}{17}+\dfrac{41}{21}\)

\(A=\dfrac{-420}{357}+\dfrac{697}{357}=\dfrac{277}{357}\)

\(B=16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)-28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)

\(B=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)-28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)\)

\(B=\left(-\dfrac{5}{3}\right)\left(16\dfrac{2}{7}-28\dfrac{2}{7}\right)\)

\(B=\left(-\dfrac{5}{3}\right)\left(-12\right)\)

\(B=20\)