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\(\frac{a+2015}{a-2015}=\frac{b+2016}{b-2016}\Rightarrow\)\(\frac{a+2015}{a-2015}-1=\frac{b+2016}{b-2016}-1\)
\(\frac{a+2015-a+2015}{a-2015}=\frac{b+2016-b+2016}{b-2016}\Rightarrow\)\(\frac{2015}{a-2015}=\frac{2016}{b-2016}\Rightarrow\)
2015(b-2016) =2016(a-2015) =>2015b =2016a =>\(\frac{a}{b}=\frac{2015}{2016}\)
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Ta có : \(\frac{a+2014}{a-2014}=\frac{a+2015}{a-2015}\)
\(\Rightarrow\left(a+2014\right)\left(a-2015\right)=\left(a-2014\right)\left(a+2015\right)\)
\(\Rightarrow a^2-a-2014.2015=a^2+a-2014.2015\)
\(\Leftrightarrow a^2-a=a^2+a\)
=> a2 - a2 - a = a
=> -a = a
=> 0 = a + a
=> 2a = 0
=> a = 0
Vậy \(\frac{a}{2014}=\frac{b}{2015}\) (đpcm)
Cho a,b,c là số dương . Chứng minh:s^2016+b^2016+c^2016>(b+c×a^2015)/2+(c+a×b^2015)/2+(a+b×a^2015)/2
Ta có A= 1/2015 + 2/2016 + 3/2017 + ... +2016/4030- 2016
A= 2015-2014/2015 + 2016-2014/2016 +...+4030-2014/4030-2016
A= 2015/2015-2014/2015+ 2016/2016-2014/2016 + ..... +4030/4030-2014/4030 -2016
A= 1-2014/2015 + 1-2014/2016 +....+1-2014/4030 -2016
A= (1+1+1+1+........+1) -(2014/2015+2014/2016+......+2014/4030) -2016
A=2016 - 2014.(1/2015+1/2016+....+1/4030) -2016
A= (2016 - 2016 ) - 2014. ( 1/2015+1/2016+.....+1/4030)
A=-2014.(1/2015+1/2016+....+1/4030)
mà B = 1/2015+1/2016+....+1/4030
nên A : B = -2014
ta có \(\frac{a+2015}{a-2015}=\frac{b+2016}{b-2016}\)
\(\Rightarrow\frac{\left(a+2015\right)-\left(a-2015\right)}{a-2015}=\frac{\left(b+2016\right)-\left(b-2016\right)}{b-2016}\)
\(\Rightarrow\frac{2015.2}{a-2015}=\frac{2016.2}{b-2016}\)
\(\Rightarrow\frac{2015}{a-2015}=\frac{2016}{b-2016}\)
\(\Rightarrow\frac{a-2015}{2015}=\frac{b-2016}{2016}\)
\(\Rightarrow\frac{a}{2015}-1=\frac{b}{2016}-1\)
Suy ra ĐPCM