a)\(\frac{2}{3}\)

b)\(\frac{3}{4}\)

\(a,\frac{1}{2}\times\frac{4}{5}\div\frac{6}{10}\)

\(=\frac{2}{5}\div\frac{6}{10}\)

\(=\frac{2}{5}\times\frac{10}{6}\)

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

\(b,\frac{24}{35}\div\left(\frac{4}{5}\times\frac{8}{7}\right)\)

\(=\frac{24}{35}\div\frac{32}{35}\)

\(=\frac{24}{35}\times\frac{35}{32}\)

\(=\frac{3}{4}\)

a=2/3 [ đã rút gọn]

b=21/28  [ đã rút gọn]

A = (4+\(\frac{1}{5}\)) . \(\frac{18}{19}\)+ (2+\(\frac{8}{5}\)) . \(\frac{21}{5}\)

A= \(\frac{21}{5}\).18/19 + 18/5 . 21/5

A= 21/5 (18/19 + 18/5)

A= 21/5 . 432/95

A= 9288/95

b= 25/2. (3+2/7) - 23/7. (5 + 1/2)

b= 25/2 . 23/7 - 23/7 . 11/2

b= 23/7 (25/2 -11/2)

b=23/7 . 7

b= 23

c=39/64

d=913/105

3) C thiếu đề

4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)

\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)

\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)

\(D=0+0+\frac{125}{14}-\frac{7}{30}\)

\(D=\frac{913}{105}\)

\(C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)

\(C=\frac{5}{18}+\frac{8}{19}-\frac{1}{3}+\frac{-5}{18}+\frac{11}{19}+\frac{1}{3}-\frac{5}{8}\)

\(C=\left(\frac{5}{18}+\frac{-5}{18}\right)+\left(\frac{8}{19}+\frac{11}{19}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)-\frac{5}{8}\)

\(C=0+1+0-\frac{5}{8}\)

\(C=\frac{3}{8}\)

Ta có:

\(\begin{array}{l}\sin \frac{\pi }{{24}}\cos \frac{{5\pi }}{{24}} = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{{24}} + \frac{{5\pi }}{{24}}} \right) + \sin \left( {\frac{\pi }{{24}} - \frac{{5\pi }}{{24}}} \right)} \right]\\ = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{4}} \right) + \sin \left( { - \frac{\pi }{6}} \right)} \right]\\ = \frac{1}{2}\left[ {\frac{{\sqrt 2 }}{2} - \frac{1}{2}} \right] = \frac{{\sqrt 2  - 1}}{4}\end{array}\)

Ta có:

\(\begin{array}{l}\sin \frac{{7\pi }}{8}\sin \frac{{5\pi }}{8} = \frac{1}{2}\left[ {\cos \left( {\frac{{7\pi }}{8} - \frac{{5\pi }}{8}} \right) - \cos \left( {\frac{{7\pi }}{8} + \frac{{5\pi }}{8}} \right)} \right]\\ = \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{{3\pi }}{2}} \right)} \right]\\ = \frac{1}{2}.\left( {\frac{{\sqrt 2 }}{2} + 0} \right) = \frac{{\sqrt 2 }}{4}\end{array}\)

a) Cách 1:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)

Cách 2:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)

b)

\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)

\(A=-1-4=-5\)

\(B=\frac{4}{3}.\frac{15}{7}-16\)

\(B=\frac{20}{7}-16\)

\(B=\frac{-92}{7}\)

\(C=\frac{28}{15}.0,25.3+\left(\frac{8}{15}-\frac{1}{4}\right)\div1\frac{23}{24}\)

\(C=1,4+\frac{17}{60}\div\frac{47}{24}\)

\(C=1,4+\frac{34}{235}\)

\(C=\frac{363}{235}\)

\(A=\frac{-15}{8}+\frac{7}{8}-4\)

\(=-1-4=-5\)

\(B=\left(4-2\frac{2}{3}\right).2\frac{1}{7}-1\frac{3}{5}:\frac{1}{10}\)

\(=\frac{4}{3}.\frac{15}{7}-\frac{8}{5}:\frac{1}{10}\)

\(=\frac{20}{7}-16=\frac{-92}{7}\)

\(C=1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-25\%\right):1\frac{23}{24}\)

\(=\frac{28}{15}.\frac{1}{4}.3+\frac{17}{60}:\frac{47}{24}\)

\(=\frac{7}{15}.3+\frac{17}{60}:\frac{47}{24}\)

\(=\frac{7}{5}+\frac{34}{235}=\frac{363}{235}\)

\(A=-1-4=-5\)

\(B=\frac{4}{3}.\frac{15}{7}-16\)

\(B=\frac{20}{7}-16\)

\(B=\frac{-92}{7}\)

\(C=\frac{28}{15}.0,25.3+\left(\frac{8}{15}-\frac{1}{4}\right):1\frac{23}{24}\)

\(C=1,4+\frac{17}{60}:\frac{47}{24}\)

\(C=1,4+\frac{34}{235}\)

\(C=\frac{363}{235}\)

a)

 \(\begin{array}{l}M = \frac{1}{7}.(\frac{{ - 5}}{8}) + \frac{1}{7}.(\frac{{ - 11}}{8})\\ = \frac{{ - 5}}{{56}} + \frac{{ - 11}}{{56}} = \frac{{ - 16}}{{56}} = \frac{{ - 2}}{7}\end{array}\)

b)

\(\begin{array}{l}M = \frac{1}{7}.(\frac{{ - 5}}{8}) + \frac{1}{7}.(\frac{{ - 11}}{8})\\ = \frac{1}{7}.[(\frac{{ - 5}}{8}) + (\frac{{ - 11}}{8})]\\ = \frac{1}{7}.\frac{{ - 16}}{8}\\ = \frac{1}{7}.( - 2)\\ = \frac{{ - 2}}{7}\end{array}\)

\(\frac{6:\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}\)

\(=\frac{\frac{6}{1}:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}\times\frac{57}{11}}\)

\(=\frac{\frac{6}{1}\times\frac{5}{3}-1}{\frac{210}{55}+\frac{57}{11}}\)

\(=\frac{\frac{30}{3}-1}{\frac{42}{11}+\frac{57}{11}}\)

\(=\frac{10-1}{\frac{99}{11}}\)

\(=\frac{9}{9}\)

\(=1\)

\(6:\frac{3}{5}-1\frac{1}{6}\)X \(\frac{6}{7}\)                                                     \(4\frac{1}{5}\)X \(\frac{10}{11}+5\frac{2}{11}\)

\(=\frac{33}{5}-\frac{7}{6}\)X \(\frac{6}{7}\)                                             \(=\)      \(\frac{21}{5}\)X  \(\frac{10}{11}+\frac{57}{11}\) 

\(=\frac{33}{5}-1\)                                                              \(=\frac{42}{11}+\frac{57}{11}\)

\(=\frac{28}{5}\)                                                                       \(=\frac{99}{11}=9\)