Gọi biểu thức đó là S
\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{5.6}\)
Ta có công thức :
\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức, ta có
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{5}-\dfrac{1}{6}\)
\(S=1-\dfrac{1}{6}=\dfrac{5}{6}\)
Ai thấy đúng thì ủng hộ nha !!!
\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)
= \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)
= 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)
= 1-\(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=1-\dfrac{1}{6}\)
\(=\dfrac{6}{6}-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
Lời giải:
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(\Rightarrow A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{5.6}\)
\(\Rightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{5}-\dfrac{1}{6}\)
\(\Rightarrow A=1-\dfrac{1}{6}\)
\(\Rightarrow A=\dfrac{5}{6}\)
