\(C=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\\ =\left(\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{301.304}\right)-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\\ =\dfrac{2}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{301}-\dfrac{1}{304}\right)-\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\\ =\dfrac{2}{3}\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\\ =\dfrac{2}{3}.\dfrac{75}{304}-\dfrac{3}{4}.\dfrac{16}{81}\\ =\dfrac{67}{4104}\)

b) Áp dụng câu a: đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\Rightarrow x+y+z=0\)

 \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\\ =x^3+y^3+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\\ =0\)

\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

a) \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

a) Ta có 

\(A=1991\times1999=\left(1995-4\right)\left(1995+4\right)\\ =1995\times1995+1995\times4-1995\times4-4\times4\\ =1995\times1995-4\times4< 1995\times1995\)

Vậy \(A< B\)

b) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}\)

\(=\dfrac{1}{3}\times\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\\ =\dfrac{1}{3}\times\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{32}\right)\\ =\dfrac{1}{3}\times\left(2-\dfrac{1}{32}\right)\\ =\dfrac{1}{3}\times\dfrac{63}{32}=\dfrac{21}{32}\)