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KN
15 tháng 4 2019
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 \(\left(2\right)\Rightarrow1< A< 2\left(đpcm\right)\)
TD
1
16 tháng 5 2017
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
HN
0
8 tháng 5 2017
Ta có :
\(A=\dfrac{50}{111}+\dfrac{50}{112}+\dfrac{50}{113}+\dfrac{50}{114}\)
Ta thấy :
\(\dfrac{50}{111}>\dfrac{50}{200}\)
\(\dfrac{50}{112}>\dfrac{50}{200}\)
\(\dfrac{50}{113}>\dfrac{50}{200}\)
\(\dfrac{50}{114}>\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}.4=1\) \(\left(1\right)\)
Mặt khác :
\(\dfrac{50}{111}< \dfrac{50}{100}\)
\(\dfrac{50}{112}< \dfrac{50}{100}\)
\(\dfrac{50}{113}< \dfrac{50}{100}\)
\(\dfrac{50}{114}< \dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}.4=2\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow1< A< 2\rightarrowđpcm\)
13 tháng 5 2018
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
NT
5 tháng 9 2016
Ta biến đổi vế phải :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\\ \)\(\\ =\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{25}\right)\\ =\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}\)
Vậy \(\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
IM
5 tháng 9 2016
Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)
=> \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\) ( đpcm )
31 tháng 3 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{60}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)
em lp 5 nen ko biet!
\(\frac{50}{111}>\frac{1}{4};\frac{50}{112}>\frac{1}{4};\frac{50}{113}>\frac{1}{4};\frac{50}{114}>\frac{1}{4}\)
\(A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}>\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)(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) \(\Rightarrow1< A< 2\)