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Ta có:
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
Đặt \(I=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
Ta có: \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};.....;\frac{9999}{10000}< \frac{10000}{10001}\)
\(\Rightarrow C< D\)
Lại có: \(C\cdot D=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\right)\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{10001}\)
Mà C<D \(\Rightarrow C\cdot C< C\cdot D\)
Hay \(C\cdot C< \frac{1}{10001}\)
\(\Rightarrow C< \frac{1}{10001}< \frac{1}{100}\)
Vậy \(C< \frac{1}{100}\left(đpcm\right)\)
Đặt :\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
\(N=\frac{2}{3}.\frac{4}{5}...\frac{10000}{10001}\)
Ta thấy:\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};....;\frac{9999}{10000}< \frac{10000}{10001}\)
Mặt khác ta thấy:
\(C.N=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\right)\)
\(C.N=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{9999}{10000}.\frac{10000}{10001}\)
\(C.N=\frac{1.2.3....9999.10000}{2.3.4....10000.10001}\)
Rút gọn phép tính \(C.N\)
\(C.N=\frac{1}{10001}\)
\(C.C< N\Rightarrow C.C< C.N\)
Hay\(C.C< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{10}.\frac{1}{10}\)
\(\Rightarrow C< \frac{1}{10000}\)(đpcm)
Đặt :
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}\)
Đặt :
B=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{9998}{9999}.\frac{10000}{10000}\)
Ta thấy " A<B
\(\Rightarrow A.A< A.B=\frac{1}{100^2}\\ \Rightarrow A^2< \frac{1}{100^2}\\ \Rightarrow A< \frac{1}{100}\)
Đặt \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}\)\(\left(A>0\right)\)
.Và \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\)\(\left(B>0\right)\)
Mặt khác :
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
... ... ...
\(\frac{9999}{10000}< \frac{10000}{10001}\)
Nhân tất cả vế theo vế \(\Rightarrow A< B\Rightarrow A^2< A.B\left(2\right)\)
(1),(2) \(\Rightarrow A^2< \frac{1}{10001}\Rightarrow A< \sqrt{\left(\frac{1}{10001}\right)}< \sqrt{\left(\frac{1}{10000}\right)}=\frac{1}{100}\left(ĐPCM\right)\)
a)đặt B=1/2.3+1/3.4+...+1/99.100
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)
từ (1),(2),(3) =>A<2
b,c tự làm
A=\(1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+..+\)\(\frac{1}{99.100}=\)\(1-\frac{1}{100}< 1\)
Mà A=1+B=>A=1+B<1+1=2
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
vậy \(A=\frac{99}{100}< 2\left(đpcm\right)\)
B)
ta có : \(1=1\)
\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{7}< \frac{1}{4}+...+\frac{1}{4}=\frac{4}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+...+\frac{1}{8}=\frac{8}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{63}< 1\)
tất cả công lại \(\Rightarrow B< 6\)
\(a)\) Đặt \(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\) ta có :
\(A=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2013+2}{2013}\)
\(A=\frac{2014}{2014}-\frac{1}{2014}+\frac{2015}{2015}-\frac{1}{2015}+\frac{2013}{2013}+\frac{2}{2013}\)
\(A=1-\frac{1}{2014}+1-\frac{1}{2015}+1+\frac{2}{2013}\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2014}+\frac{1}{2015}-\frac{2}{2013}\right)\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\left(\frac{1}{2013}+\frac{1}{2013}\right)\right]\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2013}\right]\)
\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]\)
Mà :
\(\frac{1}{2014}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2014}-\frac{1}{2013}< 0\)
\(\frac{1}{2015}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2015}-\frac{1}{2013}< 0\)
Từ (1) và (2) suy ra : \(\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)< 0\) ( cộng theo vế )
\(\Rightarrow\)\(-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>0\)
\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>3\) ( cộng hai vế cho 3 )
\(\Rightarrow\)\(A>3\) ( điều phải chứng minh )
Vậy \(A>3\)
Chúc đệ học tốt ~
c,
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)
vì \(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.............................
\(\frac{9999}{10000}< \frac{10000}{10001}\)
nên \(C^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)
\(\Rightarrow C< \frac{1}{100}\)
bt lm mỗi một câu :v
,mình sửa lại đề:
\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}< 3\)
xóa các chữ số ở tử và mẫu: 2014 và 2014,2015 và 2015
=\(\frac{2013}{2013}\)
=\(1\)
vì \(1>3\) nên \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)