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\(5B=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}\)
\(5B-B=4B=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2014}}-\frac{1}{5^{2015}}< \frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2014}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)
\(5A-A=4A=1-\frac{1}{5^{2014}}< 1\)
=>A<1/4
Ta có 4B<A<1/4
=>B<1/16( đpcm)
\(a)\) 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 \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
Ta có : \(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)
\(\Rightarrow5P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)
Lấy 5P trừ P theo vế ta có :
\(5P-P=\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\right)\)
\(\Rightarrow4P=\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt S = \(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\)
\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
Lấy 5S trừ S theo vế ta có :
\(5S-S=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\right)\)
4S = \(1-\frac{1}{5^{11}}\)
S \(=\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó : 4P = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{11}{5^{12}}\)
\(\Rightarrow P=\left(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{11}{5^{12}}\right):4=\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{11}{5^{12}.4}\right)< \frac{1}{16}\)(ĐPCM)
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
vậy 1/5.2 + 34/3456.23 =vgy0 nên ta có :
1/2.5 + B = 1/16 - B = 32156.097 : 35.98 + -9 -76 , suy ra
B= >89 _980 - -50 + 678 x 54=143.098-2014/5.2015
vậy B=78
Chua hoc
Hk tot,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhe Nguyen Chau Tuan Kiet