Question

Tính tổng A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{49.50}\)

Answer 1

Ta có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}\)

\(\Rightarrow A=\dfrac{49}{50}\)

Vậy \(A=\dfrac{49}{50}.\)

Answer 2

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=1-\dfrac{1}{50}=\dfrac{49}{50}\)

Answer 3

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Answer 4

Ta có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{49}-\dfrac{1}{49}+\dfrac{1}{50}\)

\(A=1-\dfrac{1}{50}\)

\(\Rightarrow A=\dfrac{49}{50}\)

*tích cho mình với nha*

Answer 5

A=1/1*2+1/2*3+1/3*4+...+1/49*50

=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50

=1-1/50

=0,98

Answer 6

A=1-\(\dfrac{1}{2}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)=1-\(\dfrac{1}{50}=\dfrac{49}{50}\)

mk làm vậy thôi

Answer 7

\(\dfrac{49}{50}\)

Answer 8

tự túc đi bạn ơi bài này dễ mà

Answer 9

dễ quá lớp 5 thì có