\(A=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{-5}{7}\)
\(=\dfrac{-5}{7}+\dfrac{-5}{7}+1=\dfrac{-3}{7}\)
\(B=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(=\dfrac{7}{10}.\dfrac{5}{28}.20=\dfrac{5}{2}.\)
Ta có :
A = \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
= \(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}+1\)
= \(\left(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}\right)+1\)
= \(\dfrac{5}{7}.\left(\dfrac{-2}{11}-\dfrac{9}{11}+1\right)+1\)
= \(\dfrac{5}{7}.0+1\)
= 1
B = \(0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
= \(\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
= \(\left(\dfrac{7}{10}.20\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\dfrac{5}{28}\)
= 14.1.\(\dfrac{5}{28}\)
= \(\dfrac{5}{2}\)
Vậy A = 1
B = \(\dfrac{5}{2}\)