Question

\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{108}+\frac{1}{109}\) và \(B=\frac{108}{1}+\frac{107}{2}+...+\frac{2}{107}+\frac{1}{108}\).Tính \(\frac{A}{B}\)

Answer 1

Đặt \(S=\frac{A}{B}\)

Biến đổi B 

 \(B=\frac{108}{1}+\frac{107}{2}+...+\frac{1}{108}\)

\(=\left(\frac{108}{1}+1\right)+\left(\frac{107}{2}+1\right)+...+\left(\frac{1}{108}+1\right)-108\)

\(=109+\frac{109}{2}+...+\frac{109}{108}-108\)

\(=109+\frac{109}{2}+...+\frac{109}{108}+\frac{109}{109}-109\)

\(=109.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{109}\right)\)

\(\Rightarrow s=\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{109}}{109.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{109}\right)}=\frac{1}{109}\)

KO hiểu em hỏi nhé

Answer 2

Em ko cần đặt \(S=\frac{A}{B}\)cũng được nhé tại vì anh có thói quen đặt

Answer 3

\(B=\frac{108}{1}+\frac{107}{2}+...+\frac{2}{107}+\frac{1}{108}\)

\(\Rightarrow B=1+\left(1+\frac{107}{2}\right)+...+\left(1+\frac{2}{107}\right)+\left(1+\frac{1}{108}\right)\)\(\Rightarrow B=\frac{109}{109}+\left(\frac{109}{2}\right)+...+\left(\frac{109}{107}\right)+\left(\frac{109}{108}\right)\)

\(\Rightarrow B=\left(\frac{109}{2}\right)+...+\left(\frac{109}{107}\right)+\left(\frac{109}{108}\right)+\frac{109}{109}\)

\(\Rightarrow B=109\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{108}+\frac{1}{109}\right)\)

Dấu ngoặc ở B giống A nên \(\frac{A}{B}=\frac{1}{109}\)