ta có 1+2+...+n=n*(n+1)/2

áp dụng vào A, ta được:

\(A=5+\dfrac{5}{\dfrac{2.3}{2}}+\dfrac{5}{\dfrac{3.4}{2}}+\dfrac{5}{\dfrac{4.5}{2}}+...+\dfrac{5}{\dfrac{100.101}{2}}\)

\(=5+\dfrac{10}{2.3}+\dfrac{10}{3.4}+..+\dfrac{10}{100.101}\)

\(=5+10\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\right)\)

\(=5+10\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

\(=5+10\left(\dfrac{1}{2}-\dfrac{1}{101}\right)=\dfrac{1000}{101}\)

\(\dfrac{3\times145+3\times55}{6\times215-6\times15}=\dfrac{3\times\left(145+55\right)}{6\times\left(215-15\right)}=\dfrac{3\times200}{6\times200}=\dfrac{1}{2}\)