Bài 5:
a: M là trung điểm của OA
=>\(OM=MA=\dfrac{OA}{2}=3\left(cm\right)\)
b: N là trung điểm của MA
=>\(MN=\dfrac{MA}{2}=\dfrac{3}{2}=1,5\left(cm\right)\)
bài 2:
a: \(-\dfrac{5}{8}+x=\dfrac{7}{12}\)
=>\(x=\dfrac{7}{12}+\dfrac{5}{8}\)
=>\(x=\dfrac{14}{24}+\dfrac{15}{24}=\dfrac{29}{24}\)
b: \(x-\dfrac{4}{5}=3\dfrac{1}{2}\)
=>\(x-\dfrac{4}{5}=\dfrac{7}{2}\)
=>\(x=\dfrac{7}{2}+\dfrac{4}{5}=\dfrac{35+8}{10}=\dfrac{43}{10}\)
c: \(x\cdot\dfrac{-12}{5}=\dfrac{7}{-35}\)
=>\(x\cdot\dfrac{12}{5}=\dfrac{1}{5}\)
=>\(x=\dfrac{1}{5}:\dfrac{12}{5}=\dfrac{1}{5}\cdot\dfrac{5}{12}=\dfrac{1}{12}\)
Bài 1(Đề 1):
a: \(2+\dfrac{1}{4}-\dfrac{7}{5}\)
\(=\dfrac{2\cdot20+5-7\cdot4}{20}\)
\(=\dfrac{40+5-28}{20}=\dfrac{40-23}{20}=\dfrac{17}{20}\)
b: \(-\dfrac{2}{3}\cdot\dfrac{5}{4}+\dfrac{2}{5}:\dfrac{-3}{25}\)
\(=\dfrac{-10}{12}+\dfrac{2}{5}\cdot\dfrac{-25}{3}\)
\(=-\dfrac{5}{6}+\dfrac{-50}{15}=-\dfrac{5}{6}-\dfrac{10}{3}=\dfrac{-5-20}{6}=-\dfrac{25}{6}\)
c: \(\left(2\dfrac{3}{4}-\dfrac{1}{5}\right):\dfrac{-3}{5}\cdot\dfrac{5}{9}\)
\(=\left(\dfrac{11}{4}-\dfrac{1}{5}\right)\cdot\dfrac{-5}{3}\cdot\dfrac{5}{9}\)
\(=\dfrac{55-4}{20}\cdot\dfrac{-25}{27}\)
\(=\dfrac{51}{20}\cdot\dfrac{-25}{27}=\dfrac{51}{27}\cdot\dfrac{-25}{20}=-\dfrac{5}{4}\cdot\dfrac{17}{9}=\dfrac{-85}{36}\)
Bài 1(Đề 2)
a: \(\dfrac{3}{10}-\dfrac{-2}{5}+\dfrac{7}{6}\)
\(=\dfrac{3}{10}+\dfrac{2}{5}+\dfrac{7}{6}\)
\(=\dfrac{9+12+35}{30}=\dfrac{44+12}{30}=\dfrac{56}{30}=\dfrac{28}{15}\)
b: \(-\dfrac{3}{8}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{8}{-3}+\dfrac{1}{3}:\dfrac{8}{-3}\)
\(=\dfrac{-3}{8}\cdot\dfrac{1}{2}+\dfrac{-8}{18}+\dfrac{1}{3}\cdot\dfrac{-3}{8}\)
\(=\dfrac{-3}{8}\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\dfrac{-4}{9}\)
\(=\dfrac{-3}{8}\cdot\dfrac{5}{6}+\dfrac{-4}{9}=\dfrac{-5}{16}+\dfrac{-4}{9}=\dfrac{-109}{144}\)
c: \(\left(\dfrac{7}{-3}+2\dfrac{1}{6}\right):\dfrac{-6}{5}\)
\(=\left(-\dfrac{7}{3}+2+\dfrac{1}{6}\right)\cdot\dfrac{-5}{6}\)
\(=\left(-\dfrac{1}{3}+\dfrac{1}{6}\right)\cdot\dfrac{-5}{6}=\dfrac{-1}{6}\cdot\dfrac{-5}{6}=\dfrac{5}{36}\)