a) \(\left(-5\right)^{-1}=-\dfrac{1}{5}\)

b) \(2^0\cdot\left(\dfrac{1}{2}\right)^{-5}=1\cdot32=32\)

c) \(6^{-2}\cdot\left(\dfrac{1}{3}\right)^{-3}:2^{-2}\)

\(=\dfrac{1}{36}\cdot27:\dfrac{1}{4}\)

\(=\dfrac{27\cdot4}{36}=3\)

a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)

b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)

c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)

a) \(\left(-\frac{3}{4}\right)^2:\left(\frac{5}{4}\right)^2+14,7-1\frac{9}{25}\)

\(=\left[\left(-\frac{3}{4}\right):\frac{5}{4}\right]^2+\frac{147}{10}-\frac{34}{25}\)

\(=\left[\left(-\frac{3}{4}\right)\cdot\frac{4}{5}\right]^2+\frac{147}{10}-\frac{34}{25}\)

\(=\left(-\frac{3}{5}\right)^2+\frac{147}{10}-\frac{34}{25}=\frac{9}{25}+\frac{147}{10}-\frac{34}{25}=\left(\frac{9}{25}-\frac{34}{25}\right)+\frac{147}{10}=-1+\frac{147}{10}=\frac{137}{10}\)

b) \(\left(2\frac{1}{3}-1,5\right):\left(-6\frac{1}{6}+5\frac{1}{2}\right)+2,75\)

\(=\left(\frac{7}{3}-\frac{3}{2}\right):\left(-\frac{37}{6}+\frac{11}{2}\right)+\frac{11}{4}\)

\(=\frac{5}{6}:\left(-\frac{2}{3}\right)+\frac{11}{4}=\frac{5}{6}\cdot\left(-\frac{3}{2}\right)+\frac{11}{4}=-\frac{5}{4}+\frac{11}{4}=\frac{3}{2}\)

                                                                Bài giải

\(a,\text{ }\left(-\frac{3}{4}\right)^2\text{ : }\left(\frac{5}{4}\right)^2+14,7-1\frac{9}{25}\)

\(=\frac{9}{16}\text{ : }\frac{25}{16}+\frac{147}{10}-\frac{34}{25}\)

\(=\frac{18}{50}+\frac{735}{50}-\frac{68}{50}\)

\(=\frac{685}{50}=\frac{137}{10}\)

a)

 \(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)

b)

\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)

a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)

b)

\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)

Ta có:

a) \(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3  + 3\sqrt 2 }}{6}\)      

b) \(\cos \left( {\alpha  + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} =  - \frac{{3 + \sqrt 6 }}{6}\)

c) \(\sin \left( {\alpha  - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)

d) \(\cos \left( {\alpha  - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)

ảo                         

a) \(\sqrt[4]{\dfrac{1}{16}}=\dfrac{1}{2}\)

b) \(\left(\sqrt[6]{8}\right)^2=\sqrt[\dfrac{6}{2}]{8}=\sqrt[3]{8}=2\)

c) \(\sqrt[4]{3}\cdot\sqrt[4]{27}=\sqrt[4]{3\cdot27}=\sqrt[4]{81}=3\)

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)

\(B=\frac{1}{4}\)