\(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\Rightarrow\left(a+2014\right)\left(b-2015\right)=\left(a-2014\right)\left(b+2015\right)\)

\(\Rightarrow\frac{a+2014}{b+2015}=\frac{a-2014}{b-2015}=\frac{a+2014+a-2014}{b+2015+b-2015}=\frac{2a}{2b}=\frac{a}{b}\)

\(\Rightarrow\frac{a+2014}{b+2015}=\frac{a}{b}=\frac{a+2014-a}{b+2015-b}=\frac{2014}{2015}\)

\(\frac{a}{b}=\frac{2014}{2015}\Rightarrow2015a=2014b\Rightarrow\frac{a}{2014}=\frac{b}{2015}\)

\(\Rightarrowđpcm\)

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)

Chia cả tử và mẫu của mỗi phân số tương ứng cho b²⁰¹⁵; b²⁰¹⁴

=> cần chứng minh: \(\frac{\left(\frac{a}{b}\right)^{2015}-1}{\left(\frac{a}{b}\right)^{2015}+1}>\frac{\left(\frac{a}{b}\right)^{2014}-1}{\left(\frac{a}{b}\right)^{2014}+1}\)

Ta có: \(VT=\frac{\left(\frac{a}{b}\right)^{2015}-1}{\left(\frac{a}{b}\right)^{2015}+1}=\frac{\left(\frac{a}{b}\right)^{2015}+1}{\left(\frac{a}{b}\right)^{2015}+1}-\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}=1-\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}\)

\(VP=\frac{\left(\frac{a}{b}\right)^{2014}-1}{\left(\frac{a}{b}\right)^{2014}+1}=\frac{\left(\frac{a}{b}\right)^{2014}+1}{\left(\frac{a}{b}\right)^{2014}+1}-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}=1-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}\)

Vì a> b > 0 => a/b  > 1. Do đó:

\(\left(\frac{a}{b}\right)^{2015}+1>\left(\frac{a}{b}\right)^{2014}+1\)

=> \(\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}1-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}\)

=> VT > VP 

\(\Leftrightarrow\frac{x^{2014}}{a^2+b^2+c^2+d^2}+\frac{y^{2014}}{a^2+b^2+c^2+d^2}+\frac{z^{2014}}{a^2+b^2+c^2+d^2}+\frac{t^{2014}}{a^2+b^2+c^2+d^2}\)

\(-\frac{x^{2014}}{a^2}-\frac{y^{2014}}{b^2}-\frac{z^{2014}}{c^2}-\frac{t^{2014}}{d^2}=0\)

\(\Leftrightarrow\left(\frac{x^{2014}}{a^2+b^2+c^2+d^2}-\frac{x^{2014}}{a^2}\right)+\left(\frac{y^{2014}}{a^2+b^2+c^2+d^2}-\frac{y^{2014}}{b^2}\right)+\left(\frac{z^{2014}}{a^2+b^2+c^2+d^2}-\frac{z^{2014}}{c^2}\right)\)

\(+\left(\frac{t^{2014}}{a^2+b^2+c^2+d^2}-\frac{t^{2014}}{d^2}\right)=0\)

\(\Leftrightarrow x^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+y^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\right)+\)

\(z^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\right)+t^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\right)=0\)

vì a²,b²,c²,d² lớn hơn hoặc bằng 0

=>  \(\hept{\begin{cases}\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\ne0\end{cases}}và....\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\ne0\)

\(\Rightarrow\hept{\begin{cases}x^{2014}=0\\y^{2014}=0\\z^{2014}=0\end{cases}}và..t^{2014}=0\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}và...t=0\)

=> \(\hept{\begin{cases}x^{2015}=0\\y^{2015}=0\\z^{2015}=0\end{cases}}và..t^{2015}=0\Rightarrow x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)

vậy \(x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)

A = \(\frac{2015^{2016}+1}{2015^{2015}+1}=\frac{2015^{2015}+1}{2015^{2015}+1}+\frac{2015}{2015^{2015}+1}=1+\frac{2015}{2015^{2015}+1}\)

B = \(\frac{2014^{2015}+1}{2014^{2014}+1}=\frac{2014^{2014}+1}{2014^{2014}+1}+\frac{2014}{2014^{2014}+1}=1+\frac{2014}{2014^{2014}+1}\)

Rồi bạn tự so sánh nha

CÁCH 1:

A=1và 2/2015^2014-1

B= 1và 2/2015^2014-3

Vì 1và 2/2015^2014-1 < 1và 2/2015^2014-3

Vậy A <B

CÁCH 2:

Ta biết: a/b>1=>a/b> a+n/b+n

B>1=> B= 2015^2014-1/2015^2014-3> 2015^2014-1+2/2015^2014-3+2=2015^2014+1/2015^2014-1=A

Vậy B>A