Ta có:A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)
A=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)
A=1-\(\dfrac{1}{51}=\dfrac{50}{51}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)
\(A=\dfrac{1}{1}-\dfrac{1}{51}\)
\(A=\dfrac{50}{51}\)
A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)
= \(1-\dfrac{1}{51}\)
= \(\dfrac{50}{51}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}+\dfrac{1}{50\cdot51}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)
\(A=1-\dfrac{1}{51}\)
\(A=\dfrac{51}{51}-\dfrac{1}{51}\)
\(A=\dfrac{50}{51}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)
\(A=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}+\dfrac{51-50}{50.51}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}-\dfrac{1}{51}\)
\(A=1-\dfrac{1}{51}\)
\(A=\dfrac{50}{51}\)
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)
=\(1-\dfrac{1}{51}\)
=\(\dfrac{50}{51}\)
Vậy A=\(\dfrac{50}{51}\)
