Question

Tim x biet:

(\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}\))=\(\dfrac{0,33.x}{2009}\)

Answer 1

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{97}-\dfrac{1}{100}\right)=\dfrac{0,33x}{2009}\)

\(\Leftrightarrow\dfrac{1}{3}\cdot\dfrac{99}{100}=\dfrac{0,33x}{2009}\)

\(\Leftrightarrow\dfrac{33}{100}=\dfrac{0,33x}{2009}\) <=> x = (tự tính )

Answer 2