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Ta có :
\(\frac{50}{111}>\frac{50}{200}\)
\(\frac{50}{112}>\frac{50}{200}\)
\(\frac{50}{113}>\frac{50}{200}\)
\(\frac{50}{114}>\frac{50}{200}\)
\(\Rightarrow A>\frac{50}{200}+\frac{50}{200}+\frac{50}{200}+\frac{50}{200}\)hay \(A>\frac{50}{200}.4\left(1\right)\)
Mặt khác :
\(\frac{50}{111}< \frac{50}{100}\)
\(\frac{50}{112}< \frac{50}{100}\)
\(\frac{50}{113}< \frac{50}{100}\)
\(\frac{50}{114}< \frac{50}{100}\)
\(\Rightarrow A< \frac{50}{100}+\frac{50}{100}+\frac{50}{100}+\frac{50}{100}\)hay \(A< \frac{50}{100}.4\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\Rightarrow1< A< 2\left(đpcm\right)\)
\(\frac{50}{111}>\frac{1}{4};\frac{50}{112}>\frac{1}{4};\frac{50}{113}>\frac{1}{4};\frac{50}{114}>\frac{1}{4}\)
\(A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}>\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)(1)
\(\frac{50}{111}< \frac{1}{2};\frac{50}{112}< \frac{1}{2};\frac{50}{113}< \frac{1}{2};\frac{50}{114}< \frac{1}{2}\)
\(\Rightarrow A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}< \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)(2)
từ (1) và (2) \(\Rightarrow1< A< 2\)
A<50/100+50/100+50/100+50/100=4.50/100=2
=>A<2
A>4.50/150=4/3+1+1/3>1
=>dccm
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)
\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)
\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)
\(=\frac{1}{30}\cdot\frac{31}{2}\)
\(=\frac{31}{60}\)
b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)
\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)
Mà \(\frac{3}{2}< 2\)
\(\Rightarrow1< A< 2\)
c ,Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\)
50/111 < 50/100
50/112 < 50/100
50/113 < 50/100
50/114 < 50/100
=> A < 200/100 => A < 2
50/111 > 50/200
50/112 > 50/200
50/113 > 50/200
50/114 > 50/200
=> A > 200/200 => A > 1
Vậy 1 < A < 2
AI THẤY OK ỦNG HỘ NHÉ