Question

[Mizuka]

11/1.2+11/2.3+.....+11/99.100=?

giúp mk nhé

nếu nhanh và đúng mk sẽ tk cho

Answer 1

[gàdsfàds]

=\(11\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99\cdot100}\right)\)=\(11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)=\(11\left(1-\frac{1}{100}\right)\)=11\(\frac{99}{100}\)=\(\frac{1089}{100}\)

Answer 2

[Nguyễn Thanh Hiền]

Đặt \(A=\frac{11}{1.2}+\frac{11}{2.3}+...+\frac{11}{99.100}\)

\(\Rightarrow A=11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=11\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=11.\frac{99}{100}\)

\(\Rightarrow A=\frac{1089}{100}\)