Ai giúp mk nha:
\(ChoA=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+2013}\)
CMR Nó <\(\frac{3}{4}\)
Cho \(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2013}\)
Cm nó <\(\frac{3}{4}\)Nha mn giúp mk nha còn có mỗi bài này nữa
CMR :
\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}< \frac{3}{4}\)
Giúp mk nha
Đặt A la tên của biểu thức trên
\(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}\)
\(=\frac{1}{2\left(3+1\right):2}+\frac{1}{3\left(5+1\right):2}+\frac{1}{4\left(7+1\right):2}+...+\frac{1}{1009\left(2017+1\right):2}\)
\(=\frac{2}{2.4}+\frac{2}{3.6}+\frac{2}{4.8}+....+\frac{2}{1009.2018}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{1009.1009}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}\right)\)
Ta có: \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...........
\(\frac{1}{1009^2}< \frac{1}{1008.1009}\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{1008.1009}\right)\)
\(A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}\right)\)
\(A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{1009}\right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)
Vậy ...
Đặt tổng đã cho là A
\(\frac{1}{1+3}=\frac{1}{\left(3+1\right)x2:2}=\frac{1}{2x4:2}=\frac{1}{2x4}x2=\frac{2}{2x4}\)=\(\frac{1}{2x2}\)
\(\frac{1}{1+3+5}=\frac{1}{\left(1+5\right)x3:2}=\frac{1}{3x6}x2=\frac{2}{3x6}\)=\(\frac{1}{3x3}\)
\(\frac{1}{1+3+5+....+2017}=\frac{1}{\left(1+2017\right)x1009:2}=\frac{1}{1009x2018}x2=\frac{2}{1009x2018}\)=\(\frac{1}{1009x1009}\)
Các mẫu là bạn áp dụng tính tổng đó nha ( mk làm tắt)
A=\(\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{1009x1009}\)<\(\frac{1}{2x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{1008x1009}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}\)=\(\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)
vậy A<3/4( Mk có làm tắt nên chỗ nào ko hiểu thì nhắn tin nha
\(ChoA=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+......+\frac{1}{1+3+......+2013}\)
\(Cm:a<\frac{3}{4}\)
\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+\frac{2010}{5}+....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{2014}+\frac{1}{2015}}\)
Trình bày tự luận giúp mình nha !
Khẩn cấp đó
ở tử số ta làm thế này
\(TS=\left(1+\frac{1}{2014}\right)+\left(1+\frac{1}{2013}\right)+\left(1+\frac{1}{2012}\right)+...+\left(1+\frac{2013}{2}\right)\)
\(TS=2015\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+...+\frac{1}{2}\right)\)
\(\frac{TS}{MS}=2015\)
CMR :
\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}< \frac{3}{4}\)
Giúp mk với
\(S=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+7+...+2017}\)
\(S=\frac{1}{\left[\left(1+3\right):2\right]^2}+\frac{1}{\left[\left(1+5\right):2\right]^2}+...+\frac{1}{\left[\left(2017+1\right):2\right]^2}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1008}-\frac{1}{1009}\)
\(S< \)
Còn đâu làm nốt , tao đi ngủ đây
CMR:\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2013^3}< \frac{1}{40}\)
cmr \(s< \frac{1}{3}\)biết \(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2013}{5^{2013}}+\frac{2014}{5^{2014}}\)
em thử nhân S với 5 rồi lấy 5S= S thử đi
chị làm toàn như vậy
ko bt có đc ko nữa
\(\frac{1}{3}+\frac{5}{6}:\left(X-2\frac{1}{5}\right)=\frac{3}{4}\)
Các bn giúp mk phép tìm X này nha!Ai nhanh mk cho 3 tik
\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)
<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)
<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)
<=> \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{6}\) : \(\frac{5}{12}\)
,<=> \(\left(x-2\frac{1}{5}\right)\)= 2
<=. x = 2 + \(\frac{11}{5}\)
<=> x = \(\frac{21}{5}\)
Tính A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}}\)
Ai giúp mk tick lại cho
tớ cần gấp !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!