Tính :
(\(\frac{1}{38}\)- 1).(\(\frac{1}{37}\)-1).(\(\frac{1}{36}\)-1).......(\(\frac{1}{2}\)-1)
Giúp mn nha! Mn sẽ tk cho.
Tính :
(\(\frac{1}{38}\)- 1).(\(\frac{1}{37}\)-1).(\(\frac{1}{36}\)-1).......(\(\frac{1}{2}\)-1)
Giúp mn nha! Mn sẽ tk cho.
Tính :
(\(\frac{1}{38}\)- 1).(\(\frac{1}{37}\)-1).(\(\frac{1}{36}\)-1).......(\(\frac{1}{2}\)-1)
Giúp mn nha! Mn sẽ tk cho.
Tìm x,y:
\(\frac{2}{3}\)- \(\frac{1}{3}\)( x-\(\frac{3}{2}\)) - \(\frac{1}{2}\)(2x+1) = 5
Giải nhanh giúp mn nha! Mn sẽ tk cho các bạn!!!
Bài 1.So sánh A và B biết: A=\(\frac{10^{17}+1}{10^{18}+1}\) B=\(\frac{10^{18}+1}{10^{19}+1}\)
Bài 2.So sánh S=\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)với 4
Bài 3.Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)Chứng minh rằng A<\(\frac{3}{4}\)
Bài 4.
a)Tính nhanh tổng sau:A=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
b)Tìm x biết:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1008}{2017}\)
mn giúp mk nha mk đang cần gấp
ai nhanh mk sẽ tick cho
tk mn
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
Chứng tỏ rằng:
\(T=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}< \frac{1}{2}\)
GIÚP MIK VS MN ƯI,CHI TIẾT SẼ ĐC K NHA
_ giải bừa :v _
\(T=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}\)
Ta thấy : \(\frac{1}{4^2}< \frac{1}{2.4};\frac{1}{14^2}< \frac{1}{12.14}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}< \frac{1}{2^2}+\frac{1}{2.4}+...+\frac{1}{12.14}\)
\(\Rightarrow T< \frac{1}{2^2}+\frac{1}{2}\left(\frac{2}{2.4}+...+\frac{2}{12.14}\right)\)
\(\Rightarrow T< \frac{1}{2^2}+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(\Rightarrow T< \frac{1}{4}+\frac{1}{2}.\frac{3}{7}\)
\(\Rightarrow T< \frac{13}{28}\)
Mà \(\frac{13}{28}< \frac{1}{2}\Rightarrow T< \frac{1}{2}\)
....
rút gọn biểu thức A=\(\frac{2-\frac{1}{3}+\frac{1}{4}}{2+\frac{1}{6}-\frac{1}{4}}\)
ai giải giúp mk mk sẽ tk cho nha!!! gấp
Tờ làm luôn, ko ghi đề nữa nhé
\(A=\frac{\frac{24}{12}-\frac{4}{12}+\frac{3}{12}}{\frac{24}{12}+\frac{2}{12}-\frac{3}{12}}\)
\(A=\frac{\frac{23}{12}}{\frac{23}{12}}=1\)
Vậy A=1
\(A=\frac{2-\frac{1}{3}+\frac{1}{4}}{2+\frac{1}{6}-\frac{1}{4}}\)\(=\frac{2-\frac{2}{6}+\frac{2}{8}}{2+\frac{2}{12}-\frac{2}{8}}\)\(=\frac{2\left(1-\frac{1}{6}+\frac{1}{8}\right)}{-2\left(1-\frac{1}{12}+\frac{1}{8}\right)}\)\(=-1\)
Tìm x,y:
\(\frac{2}{3}\)- \(\frac{1}{3}\)( x-\(\frac{3}{2}\)) - \(\frac{1}{2}\)(2x+1) = 5
Giải nhanh giúp mn, mn sẽ tk cho!!!❤❤❤
\(\Rightarrow\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)+\dfrac{1}{2}\left(2x+1\right)=\dfrac{-13}{3}\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}+x+\dfrac{1}{2}=\dfrac{-13}{3}\)
\(\Rightarrow\dfrac{4}{3}x=\dfrac{-13}{3}\Rightarrow x=\dfrac{-13}{4}\)
\(\frac{18}{37}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)
(( MN ƠI GIÚP MIK nha ))
(( MIK ĐANG CẦN LẮM ))
18/37 + 8/24 + 19/37 - 1/23/24 + 2/3
= 18/37 + 19/37 + 1/3 + 2/3 - 1/23/24
= 1 + 1 - 1/23/24
= 2 - 1/23/24
= 1/24
( chú ý : 1/3 là rút gọn của 8/24 )
\(\frac{18}{37}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)
\(=\frac{18}{37}+\frac{1}{3}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)
\(=\left(\frac{18}{37}+\frac{19}{37}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)-1\frac{23}{24}\)
\(=1+1-1\frac{23}{24}\)
\(=2-1\frac{23}{24}=\frac{1}{24}\)
\(\frac{18}{27}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)
\(=\frac{18}{27}+\frac{8}{24}+\frac{19}{37}-\frac{47}{24}+\frac{16}{24}\)
\(=\left(\frac{18}{37}+\frac{19}{37}\right)+\left(\frac{8}{24}+\frac{16}{24}\right)-\frac{47}{24}\)
\(=\frac{37}{37}+\frac{24}{24}-\frac{47}{24}\)
\(=1+1-\frac{47}{24}\)
\(=2-\frac{47}{24}\)
\(=\frac{48}{24}-\frac{47}{24}\)
\(=\frac{1}{24}\)
\(CMR:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}+\frac{1}{2013^2}< 1\)
Lúc này mik ghi thiếu đề, Giúp mik vs nha. Bạn nào giải đầy đủ, chi tiết và chính xác mik sẽ cho 3 tk nha ^_^
Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(.\) \(.\)
\(.\)
\(.\) \(.\)
\(.\) \(.\)
\(\frac{1}{2013^2}< \frac{1}{2012\cdot2013}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{2013^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}\)
Mà \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}=1-\frac{1}{2013}< 1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{2013^2}< 1\)
Nhớ k cho mình nhé!
Chúc các bạn học tốt!