\(A=\dfrac{-3}{5}+(\dfrac{-2}{5}-99)\\=\dfrac{-3}{5}+\dfrac{-2}{5}-99\\=(\dfrac{-3}{5}+\dfrac{-2}{5})-99\\=-1-99=-100\)
Vậy \(A=-100\)
\(B=(7\dfrac{2}{3}+2\dfrac{3}{5})-6\dfrac{2}{3}\\=7\dfrac{2}{3}+2\dfrac{3}{5}-6\dfrac{2}{3}\\=\dfrac{23}{3}+\dfrac{13}{5}-\dfrac{20}{3}\\=(\dfrac{23}{3}-\dfrac{20}{3})+\dfrac{13}{5}=1+\dfrac{13}{5}\\=\dfrac{5}{5}+\dfrac{13}{5}=\dfrac{18}{5}\)
Vậy \(B=\dfrac{18}{5}\)
