Chứng minh rằng : \(\frac{1}{4028}< \left(\frac{1}{2}.\frac{3}{4}.....\frac{2011}{2012}.\frac{2013}{2014}\right)^2< \frac{1}{2015}\)
Những câu hỏi liên quan
\(\frac{1}{4028}< \left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot.......\cdot\frac{2011}{2012}\cdot\frac{2013}{2014}\right)^2< \frac{1}{2015}\)
Chứng minh rằng
\(\frac{1}{4028}< \left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{2013}{2014}\right)^2< \frac{1}{2015}\)
Đặt: \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.....\frac{2013}{2014}\) (1)
Ta thấy \(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}.....\frac{2014}{2015}\)
Do đó nhân vế với vế, ta được:
\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2013}{2014}.\frac{2014}{2015}\)
\(\Rightarrow A^2< \frac{1}{2015}\)
Mặt khác, \(A>\frac{1}{2}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}.....\frac{2014}{2015}\) (2)
Từ (1) và (2), ta được:
\(A^2>\frac{1}{4}.\left(\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2013}{2014}.\frac{2014}{2015}\right)\)
\(\Rightarrow A^2>\frac{1}{4}.\frac{3}{2015}\Rightarrow A^2>\frac{3}{8060}>\frac{1}{4028}\)
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frac{frac{1}{2}+frac{1}{3}+......+frac{1}{2013}}{frac{2012}{1}+frac{2012}{2}+frac{2011}{3}+...+frac{1}{2013}}frac{frac{1}{2}+frac{1}{3}+..+frac{1}{2013}}{frac{2012}{1}+2+frac{2012}{2}+1+frac{2011}{3}+1+...+frac{1}{2013}+1-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{frac{2014}{1}+frac{2014}{2}+...+frac{2014}{2013}-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{2014left(1+frac{1}{2}+frac{1}{3}+...+frac{1}{2013}-1right)}frac{1}{2014}
Đọc tiếp
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
Tính nhanh
A = \(\left(\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+...+\frac{2014}{2015}\right):\left(\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}\right)\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+\frac{2010}{5}+....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{2014}+\frac{1}{2015}}\)
Trình bày tự luận giúp mình nha !
Khẩn cấp đó
ở tử số ta làm thế này
\(TS=\left(1+\frac{1}{2014}\right)+\left(1+\frac{1}{2013}\right)+\left(1+\frac{1}{2012}\right)+...+\left(1+\frac{2013}{2}\right)\)
\(TS=2015\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+...+\frac{1}{2}\right)\)
\(\frac{TS}{MS}=2015\)
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A =\(\frac{2015+\frac{2014}{2}+\frac{2013}{3}+\frac{2012}{4}+\frac{2011}{5}+.....+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2016}}=\)
tìm A
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
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Tính A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}}\)
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Tìm x biết
\(\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2014}+\frac{1}{2015}\right).x=\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}\)
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
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