a. \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3-1}{3}=\dfrac{2}{3}\); \(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5-3}{15}=\dfrac{2}{15}\)
b. Ta có \(VP=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{3}\) mà \(VP=\dfrac{2}{3}\) \(\Rightarrow VT=VP\)
Ta có \(VP=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\) mà \(VP=\dfrac{2}{3.5}=\dfrac{2}{15}\) \(\Rightarrow VT=VP\)
c. \(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}+\dfrac{2}{99.101}\)
\(=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}+\dfrac{1}{99.101}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\left(1-\dfrac{1}{101}\right)\) \(=\dfrac{200}{101}\)
a: \(\dfrac{1}{1}-\dfrac{1}{3}=1-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\)
b: \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
c: Ta có: \(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)