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b,\(\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right):2=\frac{2013}{2015}:2\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow\)\(x+1=2015\)
\(\Rightarrow x=2014\)
a, 2/3x -3/2.x-1/2x=5/12
x.(2/3-3/2-1/2)=5/12
x. -4/3=5/12
x=5/12:-4/3
x=-5/16
b,2/6+2/12+2/20+...+2/x.(x+1)=2013/2015
2/2.3+2/3.4+2/4.5+...+2/x.(x+1)=2013/2015
1/2(1-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015
1/2(1-1/x+1)=2013/2015
1-1/x+1=2013/2015 : 1/2
1-1/x+1=4206/2015
suy ra đề sai
Ta có: \(S=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)
Đặt \(M=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2019!}\)
\(\Rightarrow M< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow M< 1-\frac{1}{2019}=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)
\(\Rightarrow S< 1+\frac{2018}{2019}=\frac{2019}{2019}+\frac{2018}{2019}=\frac{4037}{2019}< 2\)
\(\Rightarrow S< 2\) ( ĐPCM )
Ta có : A= 1/2^2 +1/3^2 +....+1/2012^2 +1/2013^2
=> A= 1/2.2 +1/3.3 +....+1/2012.2012 +1/2013.2013
Do :1/2.2< 1/1.2
1/3.3 <1/2.3
.................
1/2012.2012 <1/2011.2012
1/2013.2013< 1/2012.2013
=>1/2.2 +1/3.3 +...+1/2012.2012+1/2013.2013< 1/1.2 +1/2.3+...+1/2011.2012+1/2012.2013
=>A<1/1 -1/2 +1/2 -1/3+...+1/2011-1/2012+1/2012-1/2013
=>A<1/1-1/2013
=>A<2013/2013 -1/2013
=> A< 2012/2013
Vì 2012<2013=>2012/2013<1
mà A<2012/2013=>A<1
Vậy A<1