Câu B
B=[tex]\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{85.87} =3}{\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{85.87}) =3(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{85}-\frac{1}{87}) =3(\frac{1}{3}-\frac{1}{87}) =3.\frac{28}{87} =\frac{28}{29}[/tex]
Câu C để mk nghĩ đã
