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\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\dfrac{49}{50}< 1\)
(y - \(\dfrac{1}{2}\)) : \(\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\right)\)= \(\dfrac{1}{3}\)
(y\(-\dfrac{1}{2}\)): \(\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)= \(\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\dfrac{3}{10}=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right)=\dfrac{1}{10}\)
y = \(\dfrac{3}{5}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{30}\right)+\frac{1}{12}\)
\(=\frac{11}{20}+\frac{1}{5}+\frac{1}{12}\)
\(=\frac{3}{4}+\frac{1}{12}\)
\(=\frac{10}{12}=\frac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)
sau đây là phần chữa của mình:
\(=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{3}{10}\)
= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{2}{5}\)
