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S=\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2018.2018}\)
Ta có: \(\dfrac{1}{5.5}< \dfrac{1}{4.5};\dfrac{1}{6.6}< \dfrac{1}{5.6};\dfrac{1}{7.7}< \dfrac{1}{6.7};...;\dfrac{1}{2018.2018}< \dfrac{1}{2017.2018}\)
\(\Rightarrow\) S<\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{2017.2018}\)
S<\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)
S< \(\dfrac{1}{4}-\dfrac{1}{2018}< \dfrac{1}{4}\)
\(\Rightarrow\)S<\(\dfrac{1}{4}\)
Học tốt nha
2)
S = \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\)
S = 3 . (\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\))
S = 1 . (\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{43.46}\))
S = 1 . (\(1-\dfrac{1}{4}+...+\dfrac{1}{43}-\dfrac{1}{46}\))
S = 1 . (\(1-\dfrac{1}{46}\))
S = 1 . \(\dfrac{45}{46}\)
S = \(\dfrac{45}{46}\)
=> \(\dfrac{45}{46}\) < 1
— S = 1/4 + 2/4 +...+10/4 (1)
= 1 + 1/4 + 2/4 +...+ 9/4 (2)
=> Lấy (2) trừ đi (1) ta được:
1 — 10/4 = —6/4
Vì 14 = 14/1 = 84/6 mà —6/4 < 84/6
Do đó S < 14
Ta có:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right).2n}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)\)
\(=\frac{1}{4}-\frac{1}{2n.2}\)
Vì \(\frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)
\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\)
Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\) (Đpcm)
Ta có :\(\dfrac{1}{5}< \dfrac{1}{4};\dfrac{1}{6}< \dfrac{1}{4};\dfrac{1}{7}< \dfrac{1}{4};\dfrac{1}{8}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{8}< \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{4}{4}=1\left(1\right)\)
Ta có :\(\dfrac{1}{9}< \dfrac{1}{8};\dfrac{1}{10}< \dfrac{1}{8};\dfrac{1}{11}< \dfrac{1}{8};...;\dfrac{1}{17}< \dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}< \dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}=\dfrac{8}{8}=1\left(2\right)\)
Từ (1) và (2)\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 1+1=2\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
Ta có : \(\dfrac{1}{5}=\dfrac{1}{5}\), \(\dfrac{1}{6}< \dfrac{1}{5}\), \(\dfrac{1}{7}< \dfrac{1}{5}\),...,\(\dfrac{1}{9}< \dfrac{1}{5}\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}< \dfrac{1}{5}\cdot5=1\)
\(\dfrac{1}{10}< \dfrac{1}{8},\dfrac{1}{11}< \dfrac{1}{8},...,\dfrac{1}{17}< \dfrac{1}{8}\)
Vậy \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}< \dfrac{1}{8}\cdot8=1\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{17}< 1+1=2\)
Chúc các bạn học tốt
ta có \(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}\)
\(\Rightarrow S>\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{6}{20}\)
\(\Rightarrow S>\frac{30}{20}\)
\(\Rightarrow S>1.5>1\)
\(\Rightarrow s>1\)
Ta có :
\(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}\)
\(\Rightarrow S< \frac{6}{15}+\frac{6}{15}+\frac{6}{15}+\frac{6}{15}+\frac{6}{15}\)
\(\Rightarrow S< \frac{30}{15}\)
\(\Rightarrow s< 2\)
Vậy \(1< S< 2\)
Ta có :
\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
\(S=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Nhận xét :
\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).
Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)