Đặt :
\(A=\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+..................+\dfrac{1}{10.110}\)
\(A=\dfrac{1}{100}\left(\dfrac{100}{1.101}+\dfrac{100}{2.102}+..................+\dfrac{100}{10.101}\right)\)
\(A=\dfrac{1}{100}\left(1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+..............+\dfrac{1}{10}-\dfrac{1}{101}\right)\)
\(A=\dfrac{1}{100}\left[\left(1+\dfrac{1}{2}+...........+\dfrac{1}{10}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+..........+\dfrac{1}{101}\right)\right]\)
Đặt :
\(B=\dfrac{1}{1.11}+\dfrac{1}{2.12}+...............+\dfrac{1}{100.101}\)
\(B=\dfrac{1}{10}\left(\dfrac{10}{1.11}+\dfrac{10}{2.12}+.............+\dfrac{10}{100.101}\right)\)
\(B=\dfrac{1}{10}\left(1-\dfrac{1}{11}+\dfrac{1}{2}-\dfrac{1}{12}+..............+\dfrac{1}{100}-\dfrac{1}{101}\right)\)
\(B=\dfrac{1}{10}\left[\left(1+\dfrac{1}{2}+...........+\dfrac{1}{100}\right)-\left(\dfrac{1}{11}+\dfrac{1}{12}+...............+\dfrac{1}{101}\right)\right]\)
\(=\dfrac{1}{10}\left[\left(1+\dfrac{1}{2}+.........+\dfrac{1}{10}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...........+\dfrac{1}{101}\right)\right]\)
\(\Rightarrow B=10A\)
\(\Rightarrow A.x=10A\)
\(x=10A:A\)
\(x=10\) (thỏa mãn)
Vậy \(x=10\) là giá trị cần tìm
~ Chúc bn học tốt ~
