Câu 2:
\(\dfrac{-55}{132}=\dfrac{-55:11}{132:11}=\dfrac{-5}{12}\)
Câu 3:
a: 3>-6
=>\(\dfrac{3}{14}>-\dfrac{6}{14}\)
b: \(\dfrac{7}{-12}=\dfrac{-7}{12}=\dfrac{-7\cdot3}{12\cdot3}=\dfrac{-21}{36}\)
\(\dfrac{11}{-18}=\dfrac{-11}{18}=\dfrac{-11\cdot2}{18\cdot2}=\dfrac{-22}{36}\)
mà \(-\dfrac{21}{36}>-\dfrac{22}{36}\left(-21>-22\right)\)
nên \(\dfrac{7}{-12}>\dfrac{11}{-18}\)
Câu 4:
a: \(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{15}{48}-\dfrac{10}{48}=\dfrac{5}{48}\)
b: \(\dfrac{-5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}\)
\(=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{8}{8}+\dfrac{14}{7}\)
=1+2
=3
c: \(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+...+\dfrac{2}{99\cdot100}\)
\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)
