Câu 3:
a. \(\dfrac{5}{12}+\dfrac{7}{12}\)=\(\dfrac{12}{12}=1\)
b. \(\dfrac{9}{4}-\dfrac{-7}{5}=\dfrac{45}{20}-\dfrac{-28}{20}=\dfrac{73}{20}\)
c.\(\dfrac{-5}{7}+\dfrac{3}{4}+\dfrac{-1}{5}+\dfrac{-2}{5}+\dfrac{1}{4}\)
=\(\left(\dfrac{-5}{7}+\dfrac{-2}{7}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\dfrac{-1}{5}\)
=\(\dfrac{-7}{7}+\dfrac{4}{4}+\dfrac{-1}{5}\)
= -1+ 1 +\(\dfrac{-1}{5}\)
= 0 + \(\dfrac{-1}{5}\)=\(\dfrac{-1}{5}\)
a) 5/12+ 7/12
= 12/12
=1
b) 9/4- (-7/5)
= 9/4+ 7/5
=45/ 20+ 28/ 20
= 73/ 20
a) \(\dfrac{5}{12}+\dfrac{7}{12}=\dfrac{5+7}{12}=\dfrac{12}{12}=1\)
b) \(\dfrac{9}{4}-\dfrac{-7}{5}=\dfrac{73}{20}\)
c) \(\dfrac{-5}{7}+\dfrac{3}{4}+\dfrac{-1}{5}+\dfrac{-2}{7}+\dfrac{1}{4}\) \(=\left(\dfrac{-5}{7}+\dfrac{-2}{7}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\dfrac{-1}{5}\) \(=\left(-1\right)+1+\dfrac{-1}{5}=\dfrac{-1}{5}\)