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Ta có: \(\left(2008-\dfrac{2}{135}+\dfrac{1}{50}\right)-\left(1-\dfrac{7}{135}+\dfrac{4}{150}\right)-\left(5+\dfrac{5}{135}+\dfrac{3}{50}\right)\)
= \(2008-\dfrac{2}{135}+\dfrac{1}{50}-1+\dfrac{7}{135}-\dfrac{4}{150}-5-\dfrac{5}{135}-\dfrac{3}{50}\)
= (2008-1-5) + \(\left(\dfrac{1}{50}-\dfrac{3}{50}\right)-\left(\dfrac{2}{135}-\dfrac{7}{135}\right)-\dfrac{4}{150}\)
=2002 \(-\dfrac{1}{25}\)+\(\dfrac{1}{27}\)\(-\dfrac{4}{150}\)
=2001,9(3)
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
\(\left(2018-\frac{2}{135}+\frac{1}{50}\right)-\left(1-\frac{7}{135}+\frac{4}{50}\right)-\left(5+\frac{5}{135}+\frac{3}{50}\right)\)
\(=2018-\frac{2}{135}+\frac{1}{50}-1+\frac{7}{135}-\frac{4}{50}-5-\frac{5}{135}-\frac{3}{50}\)
\(=2012-\frac{6}{50}\)