Question

4. tính

a. A=\(\dfrac{1}{3}\).\(\dfrac{4}{5}\)+\(\dfrac{1}{3}\).\(\dfrac{6}{5}\)+\(\dfrac{2}{3}\)

b. B=\(\dfrac{-5}{6}\).\(\dfrac{4}{19}\)+\(\dfrac{-7}{12}\).\(\dfrac{4}{19}\)-\(\dfrac{40}{57}\)

c. C=\(\dfrac{3}{7}\).\(\dfrac{9}{26}\)-\(\dfrac{1}{14}\).\(\dfrac{1}{13}\)-\(\dfrac{1}{7}\)

Answer 1

\(A=\dfrac{1}{3}\cdot\dfrac{4}{5}+\dfrac{1}{3}\cdot\dfrac{6}{5}+\dfrac{2}{3}\\ =\dfrac{1}{3}\cdot\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\\ =\dfrac{1\cdot2}{3}+\dfrac{2}{3}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{4}{3}\)

Answer 2

B =\(\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}\)-\(\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\left[\dfrac{-5}{6}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\left[\dfrac{-10}{12}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)

=\(\dfrac{-17}{57}\)-\(\dfrac{40}{57}\)

=\(\dfrac{-17}{57}+\dfrac{-40}{57}\)

=\(\dfrac{-57}{57}=-1\)