Vì \(\left(\right. - 0 , 25 \left.\right)^{5} : x = \left(\right. - 0 , 25 \left.\right)^{3}\) nên \(x = \left(\right. - 0 , 25 \left.\right)^{5} : \left(\right. - 0 , 25 \left.\right)^{3}\)
\(x=\left(\right.-0,25\left.\right)^2=\left(-\frac{1}{4}\right)^2=\frac{1}{16}\)

a) \(\frac{3}{4}+\frac{9}{5}\left(\frac{3}{2}-\frac{2}{3}\right)^2=\frac{3}{4}+\frac{9}{5}\left(\frac{5}{6}\right)^2=\frac{3}{4}+\frac{9}{5}\cdot\frac{25}{36}=\frac{3}{4}+\frac{5}{4}=2\)

b) \(\frac{- 22}{25}+\left(\right.\frac{22}{7}-0,12\left.\right)=\frac{- 22}{25}+\left(\right.\frac{22}{7}-\frac{12}{100}\left.\right)=\frac{- 88}{100}+\frac{22}{7}+\frac{- 12}{100};=\left(\right.\frac{- 88}{100}+\frac{- 12}{100}\left.\right)+\frac{22}{7}=-1+\frac{22}{7}=\frac{15}{7}\)

a) \(\frac{3}{4}+\frac{9}{5}\left(\frac{3}{2}-\frac{2}{3}\right)^2=\frac{3}{4}+\frac{9}{5}\left(\frac{5}{6}\right)^2=\frac{3}{4}+\frac{9}{5}\cdot\frac{25}{36}=\frac{3}{4}+\frac{5}{4}=2\)

b) \(\frac{- 22}{25}+\left(\right.\frac{22}{7}-0,12\left.\right)=\frac{- 22}{25}+\left(\right.\frac{22}{7}-\frac{12}{100}\left.\right)=\frac{- 88}{100}+\frac{22}{7}+\frac{- 12}{100};=\left(\right.\frac{- 88}{100}+\frac{- 12}{100}\left.\right)+\frac{22}{7}=-1+\frac{22}{7}=\frac{15}{7}\)

a) \(\frac{3}{4}+\frac{9}{5}\left(\frac{3}{2}-\frac{2}{3}\right)^2=\frac{3}{4}+\frac{9}{5}\left(\frac{5}{6}\right)^2=\frac{3}{4}+\frac{9}{5}\cdot\frac{25}{36}=\frac{3}{4}+\frac{5}{4}=2\)

b) \(\frac{- 22}{25}+\left(\right.\frac{22}{7}-0,12\left.\right)=\frac{- 22}{25}+\left(\right.\frac{22}{7}-\frac{12}{100}\left.\right)=\frac{- 88}{100}+\frac{22}{7}+\frac{- 12}{100};=\left(\right.\frac{- 88}{100}+\frac{- 12}{100}\left.\right)+\frac{22}{7}=-1+\frac{22}{7}=\frac{15}{7}\)

a) \(\frac{3}{4}+\frac{9}{5}\left(\frac{3}{2}-\frac{2}{3}\right)^2=\frac{3}{4}+\frac{9}{5}\left(\frac{5}{6}\right)^2=\frac{3}{4}+\frac{9}{5}\cdot\frac{25}{36}=\frac{3}{4}+\frac{5}{4}=2\)

b) \(\frac{- 22}{25}+\left(\right.\frac{22}{7}-0,12\left.\right)=\frac{- 22}{25}+\left(\right.\frac{22}{7}-\frac{12}{100}\left.\right)=\frac{- 88}{100}+\frac{22}{7}+\frac{- 12}{100};=\left(\right.\frac{- 88}{100}+\frac{- 12}{100}\left.\right)+\frac{22}{7}=-1+\frac{22}{7}=\frac{15}{7}\)

a) \(\frac{3}{4}+\frac{9}{5}\left(\frac{3}{2}-\frac{2}{3}\right)^2=\frac{3}{4}+\frac{9}{5}\left(\frac{5}{6}\right)^2=\frac{3}{4}+\frac{9}{5}\cdot\frac{25}{36}=\frac{3}{4}+\frac{5}{4}=2\)

b) \(\frac{- 22}{25}+\left(\right.\frac{22}{7}-0,12\left.\right)=\frac{- 22}{25}+\left(\right.\frac{22}{7}-\frac{12}{100}\left.\right)=\frac{- 88}{100}+\frac{22}{7}+\frac{- 12}{100};=\left(\right.\frac{- 88}{100}+\frac{- 12}{100}\left.\right)+\frac{22}{7}=-1+\frac{22}{7}=\frac{15}{7}\)