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ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{4}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2015^2}< \dfrac{1}{2014.2015}\)
=> A < \(\dfrac{1}{4}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\)
=> A< \(\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
<=> A< \(\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{2015}\right)\) = \(\dfrac{3}{4}-\dfrac{1}{2015}\) < \(\dfrac{3}{4}\).
=> đpcm.
Bài 2:
\(E=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(\Leftrightarrow E=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)
\(\Leftrightarrow E=\dfrac{3.4.5...100}{2.3.4...99}\)
\(\Leftrightarrow E=\dfrac{\left(3.4.5...99\right).100}{2.\left(3.4...99\right)}\)
\(\Leftrightarrow E=\dfrac{100}{2}=50\)
Vậy ...