giúp mình với
cho A=1/2-1/3+1/4-1/5+...+1/48+1/49
chứng minh 1/5<A<2/5
XIN CẢM ƠN
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 6 2021
a, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )
\(\Rightarrow a^2+2ab+b^2=\left(a+b\right)^2\ge4ab\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
b, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )
\(\Rightarrow ab\le\dfrac{a^2+b^2}{2}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
2 tháng 10 2025
a: \(A=1+\frac15+\frac{1}{25}+\cdots+\frac{1}{78125}\)
=>\(A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^7}\)
=>\(5\times A=5+1+\frac15+\cdots+\frac{1}{5^6}\)
=>\(5\times A-A=5+1+\frac15+\cdots+\frac{1}{5^6}-1-\frac15-\cdots-\frac{1}{5^7}\)
=>\(4\times A=5-\frac{1}{5^7}=\frac{5^8-1}{5^7}\)
=>\(A=\frac{5^8-1}{4\times5^7}\)
b:Sửa đề: \(B=\frac13+\frac{1}{12}+\cdots+\frac{1}{49152}\)
=>\(B=\frac13+\frac{1}{3\times4}+\frac{1}{3\times4^2}+\cdots+\frac{1}{3\times4^7}\)
=>\(4\times B=\frac43+\frac13+\frac{1}{3\times4}+\cdots+\frac{1}{3\times4^6}\)
=>\(4\times B-B=\frac43+\frac13+\frac{1}{3\times4}+\cdots+\frac{1}{3\times4^6}-\frac13-\frac{1}{3\times4}-\frac{1}{3\times4^2}-\cdots-\frac{1}{3\times4^7}\)
=>\(3\times B=\frac43-\frac{1}{3\times4^7}=\frac{4^8-1}{3\times4^7}\)
=>\(B=\frac{4^8-1}{9\times4^7}\)
c: \(C=\frac53+\frac56+\frac{5}{12}+\frac{5}{24}+\cdots+\frac{5}{192}+\frac{5}{384}\)
=>\(2\times C=\frac{10}{3}+\frac53+\frac56+\cdots+\frac{5}{96}+\frac{5}{192}\)
=>\(2\times C-C=\frac{10}{3}+\frac53+\frac56+\cdots+\frac{5}{96}+\frac{5}{192}-\frac53-\frac56-\cdots-\frac{5}{192}-\frac{5}{384}\)
=>\(C=\frac{10}{3}-\frac{5}{384}=\frac{1280}{384}-\frac{5}{384}=\frac{1275}{384}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
16 tháng 1 2023
a: ĐKXĐ: x>0; x<>9
b: \(A=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{x-9}:\dfrac{\sqrt{x}+3-3}{\sqrt{x}+3}\)
\(=\dfrac{2x}{x-9}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
c: Để A=-1 thì 2 căn x=-căn x+3
=>x=1
18 tháng 4 2019
ta có:
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\),
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}...\)
\(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)
Từ trên => A < \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
=> \(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{9}-\frac{1}{10}\)
=> \(A< \frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
=> \(A< \frac{2}{5}\)mà \(\frac{2}{5}< \frac{1}{2}\)
=> \(A< \frac{1}{2}\)=> \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{2}\)
Chúc bn học tốt !
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}\)
\(\Rightarrow1-A-\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}\)
\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{25}\)
\(\Rightarrow\frac{49}{50}-A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)\)
Ta có :
\(\frac{1}{26}< \frac{1}{25};\frac{1}{27}< \frac{1}{25};\frac{1}{28}< \frac{1}{25};\frac{1}{29}< \frac{1}{25};\frac{1}{30}< \frac{1}{25};\)
\(\frac{1}{31}< \frac{1}{30};\frac{1}{32}< \frac{1}{30};..;\frac{1}{39}< \frac{1}{30};\frac{1}{40}< \frac{1}{30};\)
\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{49}< \frac{1}{40};\frac{1}{50}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)>\frac{49}{50}-\frac{4}{5}=\frac{9}{50}>\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow A>\frac{1}{5}\)( đpcm )