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Vì 18/91 < 18/90 =1/5
23/114>23115=1/5
vậy 18/91<1/5<23/114
suy ra 18/91<23/114
vì 21/52=210/520
Mà 210/520=1-310/520
213/523=1-310/523
310/520>310/523
vậy 210/520<213/523
suy ra 21/52<213/523
a) Giải
Ta có: \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}+\dfrac{1}{2^{2013}}\)
\(\Rightarrow2S=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{2012}}+\dfrac{2}{2^{2013}}\)
\(2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2S-S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2012}}-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=1-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=\dfrac{2^{2013}-1}{2^{2013}}\)
b) Giải
Từ \(A=\dfrac{2011^{2012}+1}{2011^{2013}+1}\)
\(\Rightarrow2011A=\dfrac{2011^{2013}+20111}{2011^{2013}+1}=\dfrac{2011^{2013}+1+2010}{2011^{2013}+1}=1+\dfrac{2010}{2011^{2013}+1}\)
Từ \(B=\dfrac{2011^{2013}+1}{2011^{2014}+1}\)
\(\Rightarrow2011B=\dfrac{2011^{2014}+2011}{2011^{2014}+1}=\dfrac{2011^{2014}+1+2010}{2011^{2014}+1}=1+\dfrac{2010}{2011^{2014}+1}\)
Vì 20112013 + 1 < 20112014 + 1 và 2010 > 0
\(\Rightarrow\dfrac{2010}{2011^{2013}+1}>\dfrac{2010}{2011^{2014}+1}\)
\(\Rightarrow2011A>2011B\)
\(\Rightarrow A>B\)
Vậy A > B.
\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{49}{50}\)
\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{43\cdot46}\\ =\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\\ =1-\dfrac{1}{46}\\ =\dfrac{45}{46}\)
\(S=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\\ =\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{49\cdot51}\\ =\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\\ =\dfrac{1}{2}\cdot\dfrac{16}{51}\\ =\dfrac{8}{51}\)
\(S=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{98\cdot99\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{98\cdot99\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}-\dfrac{1}{99\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{99\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\\ =\dfrac{1}{2}\cdot\dfrac{4949}{9900}=\dfrac{4949}{19800}\)
Ta có S = 1/11+1/12+1/13+...+1/19+1/20 nên S có 10 số hạng
Và 1/2 = 10/20
Mà 1/11 > 1/12 > 1/13 > 1/14 > 1/15 > 1/16 > 1/17 > 1/18 > 1/19 > 1/20
Nên 1/11+1/12+1/13+...+1/19+1/20 > 1/20x10
=> 1/11+1/12+1/13+...+1/19+1/20 > 10/20
=> 1/11+1/12+1/13+...+1/19+1/20 > 1/2
Vậy S > 1/2
Ta có:5/20>5/25
5/21>5/25
5/22>5/25
5/23>5/25
5/24>5/25
=>S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=1
=>5/20+5/21+5/22+5/23+5/24>1
DỄ
DO: 5/20 <1
5/21<1
5/22<1
5/23<1
5/24<1
=> 5/20+5/21+5/22+5/23+5/24<1
hay S<1 ( ĐPCM)
ĐÚNG NÈ ỦNG HỘ
Giải:
Vì:
\(\dfrac{1}{2}< \dfrac{2}{3}.\)
\(\dfrac{2}{3}< \dfrac{3}{4}.\)
.............
\(\dfrac{9999}{10000}< \dfrac{10000}{10001}.\)
\(\Rightarrow S^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9999}{10000}.\dfrac{10000}{10001}.\)
\(\Rightarrow S^2< \dfrac{1}{10001}< \dfrac{1}{10000}=\left(\dfrac{1}{100}\right)^2.\)
Mà \(\left(\dfrac{1}{100}\right)^2=0,01^2.\)
\(\Rightarrow S^2< 0,01^2\left(=\left(\dfrac{1}{100}\right)^2\right).\)
\(\Rightarrow S< 0,01.\)
Vậy \(S< 0,01.\)
~ Học tốt!!! ~
a: 7/30=21/90
8/45=16/90
11/90=11/90
b: -4/5=-168/210
1/6=35/210
-9/7=-270/210
c: -7/24=-21/72
11/12=66/72
-23/36=-46/72
d: 17/30=85/150
-22/75=-44/150
5=750/150
=1/1.2+1/1.2.3+.............1/1.2.3.4. ......... .2019
=1-1/2+1/2-1/3+1/3-.............-2019
=1-1/2019
=2018/2019
Vay 1/1.2+1/1.2.3+.............1/1.2.3.4. ......... .2019>1/2
\(S=\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\)
\(S>\dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)