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Ta có:

\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)

\(\Rightarrow2S=2\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\right)\)

\(\Rightarrow2S=6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)

\(\Rightarrow2S-S=\left(6+3+\dfrac{3}{2}+...+\dfrac{3}{2^8}\right)-\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\right)\)

\(\Rightarrow S=6-\dfrac{3}{2^9}=6-\dfrac{3}{512}=\dfrac{3069}{512}\)

Vậy \(S=\dfrac{3069}{512}\)

Giải:

\(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}\)

Đặt \(B=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{10000}{10001}\)

Do \(\dfrac{1}{2}< \dfrac{2}{3};\dfrac{3}{4}< \dfrac{4}{5};...;\dfrac{9999}{10000}< \dfrac{10000}{10001}\)

Nên \(C< B\) Mà \(\left\{{}\begin{matrix}C>0\\B>0\end{matrix}\right.\)

\(\Rightarrow C^2< C.B=\left(\dfrac{1}{2}.\dfrac{3}{4}...\dfrac{9999}{10000}\right)\)\(\left(\dfrac{2}{3}.\dfrac{4}{5}...\dfrac{10000}{10001}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{9999}{10000}.\dfrac{10000}{10001}\)

\(=\dfrac{1.2.3.4.5.6...9999.10000}{2.3.4.5.6.7....10000.10001}\)

\(=\dfrac{1}{10001}< \dfrac{1}{10000}=\dfrac{1}{100}.\dfrac{1}{100}=\left(\dfrac{1}{100}\right)^2\)

\(\Rightarrow C^2< \left(\dfrac{1}{100}\right)^2\Leftrightarrow C< \dfrac{1}{100}\)

Vậy \(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{1}{100}\) (Đpcm)

Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+\dfrac{1}{44}+...+\dfrac{1}{80}\)

\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}\right)\)

Nhận xét:

\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\) \(=\dfrac{1}{3}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}>\dfrac{1}{80}+\dfrac{1}{80}+...+\dfrac{1}{80}\) \(=\dfrac{1}{4}\)

\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}>\dfrac{1}{12}\)

Vậy \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{12}\) (Đpcm)

Ta có:

Muốn chứng minh \(A=999993^{1999}-555557^{1997}⋮5\) ta xét chữ số tận cùng của số hạng:

\(*)\) \(999993^{1999}=\left(...3\right)^{1999}\Rightarrow\) Ta xét \(3^{1999}\)

Ta có: \(3^{1999}=\left(3^4\right)^{499}.3^3=\left(...1\right)^{499}.27=\left(...7\right)\)

\(*)\) \(555557^{1997}=\left(...7\right)^{1997}\Rightarrow\) Ta xét \(7^{1997}\)

Ta có: \(7^{1997}=\left(7^4\right)^{499}.7=\left(...1\right)^{499}.7=\left(...7\right)\)

\(\Rightarrow A=999993^{1999}-555557^{1997}=\left(...7\right)-\left(...7\right)=0\)

Mà số có chữ số tận cùng là \(0\Leftrightarrow\) Số đó chia hết cho \(5\)

Vậy \(A=999993^{1999}-555557^{1997}⋮5\) (Đpcm)