\(\dfrac{28}{5}\)x - \(\dfrac{23}{6}\)= \(\dfrac{1}{6}\)

\(\dfrac{28}{5}\)x        = \(\dfrac{1}{6}\) + \(\dfrac{23}{6}\)

\(\dfrac{28}{5}\)x        = 4

     x        = 4 : \(\dfrac{28}{5}\)

    x        =   \(\dfrac{5}{7}\)

\(-\dfrac{5}{2}\) <  \(\dfrac{3}{x}\) < \(\dfrac{2}{-3}\)

\(-\dfrac{30}{12}\) < \(\dfrac{30}{10x}\) < \(\dfrac{30}{-45}\)

-45 < 10x < -12

- \(\dfrac{9}{2}\) < x < - \(\dfrac{6}{5}\)

-4,5 < x < - 1,2

x ϵ { -4; -3; -2 }

\(\dfrac{4}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{2}{5}\cdot\dfrac{-3}{7}=\dfrac{15}{4}\)

\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{-6}{35}=\dfrac{15}{4}\)

\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)+\dfrac{6}{35}=\dfrac{15}{4}\)

\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)=\dfrac{15}{4}-\dfrac{6}{35}=\dfrac{501}{140}\)

\(\Rightarrow\dfrac{2}{3}x-\dfrac{9}{5}=\dfrac{501}{140}:\dfrac{2}{5}=\dfrac{501\cdot5}{28\cdot5\cdot2}=\dfrac{501}{56}\)

\(\Rightarrow\dfrac{2}{3}x=\dfrac{501}{56}+\dfrac{9}{5}=\dfrac{2001}{280}\)

\(\Rightarrow x=\dfrac{2001}{280}:\dfrac{2}{3}=\dfrac{2001\cdot3}{280\cdot2}=\dfrac{6003}{560}\)

Ta dùng thuật toán Euclid

1024:580 dư 444

=> ƯCLN(1024; 580)=ƯCLN(580; 444)

580:444 dư 136

=> ƯCLN(580; 444)=ƯCLN(444; 136)

444:136 dư 36

=> ƯCLN(444; 136)=ƯCLN(136; 36)

136:36 dư 28

=> ƯCLN(136; 36)=ƯCLN(36;28)

Ta có: 36=2².3²

28=2².7

=> ƯCLN(36; 28)=ƯCLN(1024; 580)=2²=4

Làm theo cách làm tương tự, ta có ƯCLN(690; 960)=60

\(\dfrac{-5}{12}.\dfrac{2}{11}+\dfrac{-5}{12}.\dfrac{9}{11}+\dfrac{5}{12}\)

\(=\dfrac{-5}{12}\left(\dfrac{2}{11}+\dfrac{9}{11}-1\right)\)

\(=\dfrac{-5}{12}\left(1-1\right)\)

\(=0\)

\(=\left(-67\right).\left(-33\right)-67.77+\left(-67\right).33-77.33\)

\(=\left[\left(-67\right).\left(-33\right)+\left(-67\right).33\right]-66.77-77.33\)

\(=-77\left(66+33\right)=-77.99=-7623\)

\(B=\dfrac{2^{19}\cdot27^3+15.4^9\cdot9^4}{6^9+2^{10}\cdot3^{10}}\)

\(\Rightarrow B=\dfrac{2^{19}\cdot\left(3^3\right)^3+3\cdot5\cdot4^9\cdot\left(3^2\right)^4}{2^3\cdot3^3+\left(2\cdot3\right)^{10}}\)

\(\Rightarrow B=\dfrac{2^{19}\cdot3^9+3\cdot5\cdot4^9\cdot3^8}{\left(2\cdot3\right)^3+\left(2\cdot3\right)^{10}}\)

\(\Rightarrow B=\dfrac{2^{19}\cdot3^9+5\cdot\left(2^2\right)^9\cdot3^9}{\left(2\cdot3\right)^{13}}\)

\(\Rightarrow B=\dfrac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3^9}{2^{13}\cdot3^{13}}\)

\(\Rightarrow B=\dfrac{2\cdot\left(2^{18}\cdot3^9\right)+5\cdot\left(2^{18}\cdot3^9\right)}{2^{13}\cdot3^{13}}\)

\(\Rightarrow B=\dfrac{\left(2^{18}\cdot3^9\right)\cdot\left(2+5\right)}{2^{13}\cdot3^{13}}\)

\(\Rightarrow B=\dfrac{2^{18}\cdot3^9\cdot7}{2^{13}\cdot3^{13}}=\dfrac{2^{13}\cdot2^5\cdot3^9\cdot7}{2^{13}\cdot3^9\cdot3^4}\)

\(\Rightarrow B=\dfrac{2^5\cdot7}{3^4}\)