Question

Cho

\(A=\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+...+\dfrac{1}{25.125}\)

\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+\dfrac{1}{3.28}+...+\dfrac{1}{100.125}\)

Trong đó A có 25 số hạng,B có 100 số hạng.Tìm thương A:B

Answer 1

Gọi x là thương A:B cần tìm.Theo đề, ta có:

\(\left(\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{100.125}\right)x=\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{25.125}\)

Nhân 2 vế cho 100, ta có:

\(4\left(\dfrac{25}{1.26}+\dfrac{25}{2.27}+...+\dfrac{25}{100.125}\right)x=\dfrac{100}{1.101}+\dfrac{100}{2.102}+...+\dfrac{100}{25.125}\)

\(\Rightarrow4\left(1-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{100}-\dfrac{1}{125}\right)x=1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+...+\dfrac{1}{25}-\dfrac{1}{125}\)

\(\Rightarrow4\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{125}\right)\right]x=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)\)\(\Rightarrow4x=1\Rightarrow x=\dfrac{1}{4}\)

Vậy hiệu A:B là:\(\dfrac{1}{4}\)