\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(=\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{50}{49.50}-\dfrac{49}{49.50}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}=\dfrac{49}{50}\)
Vậy A = \(\dfrac{49}{50}\)
A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
A = \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{49}-\dfrac{1}{50}\)
A = \(1-\dfrac{1}{50}\)
A = \(\dfrac{49}{50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
Ta thấy :
\(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};\)
\(...;\dfrac{1}{49.50}=\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\)
\(\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A=1-\dfrac{1}{50}=\dfrac{49}{50}\)
Vậy A=\(\dfrac{49}{50}\)
A=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
A=1-1/50
(thông cảm mk ko ghi lại đề khi làm bài bn nhớ khi lại đề rồi ms giải nha
