\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}=\frac{\frac{2^3}{3^3}.\frac{3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{\left(-5\right)^3}{12^3}}=\)\(\frac{\frac{1}{6}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{5^3}{2^6.3^3}.\left(-1\right)}=\frac{\frac{1}{2.3}}{\frac{5}{2^4.3^3}}=\frac{2^3.3^2}{5}=\frac{72}{5}\)

\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)

\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)

\(=\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot12^3}}=\dfrac{-1}{6}:\dfrac{-5}{432}=\dfrac{1}{6}\cdot\dfrac{432}{5}=\dfrac{72}{5}\)

????

\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)

\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)

=>83/12x=5/12

hay x=5/83

a) $A=\left[\frac{2}{7}\left(\frac{1}{4}-\frac{1}{3}\right)\right]:\left[\frac{2}{7}\left(\frac{1}{3}-\frac{2}{5}\right)\right]=\left(\frac{1}{4}-\frac{1}{3}\right):\left(\frac{1}{3}-\frac{2}{5}\right)=1\frac{1}{4}$.
b) $B=\frac{\frac{3}{4}\left(\frac{1}{5}-\frac{2}{7}-\frac{1}{3}+\frac{2}{7}\right)}{\frac{1}{5}\left(\frac{2}{7}+\frac{1}{3}\right)-\frac{1}{3}\left(\frac{2}{7}+\frac{1}{3}\right)}=\frac{\frac{3}{4}\left(\frac{1}{5}-\frac{1}{3}\right)}{\left(\frac{1}{5}-\frac{1}{3}\right)\left(\frac{2}{7}+\frac{1}{3}\right)}=1\frac{11}{52}$.

a) $\mathrm{A}=\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right)\right]:\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)\right]=\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right):\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)=1\genfrac{}{}{0ex}{}{1}{4}$.
b) $\mathrm{B}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{2}{7}-\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{2}{7}\right)}{\genfrac{}{}{0ex}{}{1}{5}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)-\genfrac{}{}{0ex}{}{1}{3}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=1\genfrac{}{}{0ex}{}{11}{52}$

Biến đổi thừa số tổng quát: \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).

Do đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\), \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\), \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\),..., \(1+\dfrac{1}{2018.2020}=\dfrac{2019^2}{2018.2020}\), \(1+\dfrac{1}{2019.2021}=\dfrac{2020^2}{2019.2021}\). Từ đó suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)\) 

\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}...\dfrac{2019^2}{2018.2020}.\dfrac{2020^2}{2019.2021}\)

\(=\dfrac{2.2020}{2021}=\dfrac{4040}{2021}\)

1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)

2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)