a.\(\dfrac{19\left(3-4+1\right)}{1995.1996.1997}\)=\(\dfrac{19.0}{1995.1996.1997}\)=0
\(A=\dfrac{19.3-19.4+19}{1995.1996.1997}=\dfrac{19.3-19.4+19.1}{1995.1996.1997}=\dfrac{19.0}{1995.1996.1997}=\dfrac{0}{1995.1996.1997}=0\)b) sửa đề:
\(B=\dfrac{48.48-17}{47.48+31}=\dfrac{48.\left(47+1\right)-17}{47.48+31}=\dfrac{48.47+48-17}{47.48+31}=\dfrac{47.48+31}{47.48+31}=1\)
mk chỉ làm được câu a thôi
tick cho mk nhé