Question

Tính giá trị biểu thức A = \(\frac{1}{1\times2}+\frac{1}{2\times3}+...\frac{1}{99\times100}\)

Answer 1

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Answer 2

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

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

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

Answer 3

                                          Bài làm :

Ta có :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)

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

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 4

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{100-99}{99.100}\)

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

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

Answer 5

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

\(A=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...+\frac{100-99}{99.100}\)

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

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

Answer 6

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

\(A=1-\frac{1}{100}=\frac{99}{100}\)

cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt^^

Answer 7

     \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)