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Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Vì \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
=> \(\dfrac{\left(a+b\right)^{2014}}{\left(c+d\right)^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}\)
Mà \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
=> \(\dfrac{\left(a+b\right)^{2014}}{\left(c+d\right)^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}=\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}\) (1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\) (2)
Từ (1);(2) => \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+....+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{3}{2014}+1\right)+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)\)
\(B=\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+....+\dfrac{2017}{2014}+\dfrac{2017}{2015}+\dfrac{2017}{2016}\)
\(B=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
\(\dfrac{B}{A}=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}=2017\)
\(\dfrac{B}{A}=\dfrac{\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\left(\dfrac{2015}{2}+\dfrac{2}{2}\right)+\left(\dfrac{2014}{3}+\dfrac{3}{3}\right)+...+\left(\dfrac{1}{2016}+\dfrac{2016}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Vậy \(\dfrac{B}{A}=2017\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\Rightarrow\left(a+2014\right)\left(b-2015\right)=\left(a-2014\right)\left(b+2015\right)\)
\(\Rightarrow\frac{a+2014}{b+2015}=\frac{a-2014}{b-2015}=\frac{a+2014+a-2014}{b+2015+b-2015}=\frac{2a}{2b}=\frac{a}{b}\)
\(\Rightarrow\frac{a+2014}{b+2015}=\frac{a}{b}=\frac{a+2014-a}{b+2015-b}=\frac{2014}{2015}\)
\(\frac{a}{b}=\frac{2014}{2015}\Rightarrow2015a=2014b\Rightarrow\frac{a}{2014}=\frac{b}{2015}\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\\ \dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}\left(k^{2014}+1\right)}{d^{2014}\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}\\ \left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{b^{2014}}{d^{2014}}\\ \RightarrowĐPCM\)
\(\dfrac{a+2014}{a-2014}=\dfrac{b+2015}{b-2015}=\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}\) (1)
Từ (1) \(\Rightarrow\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}=\dfrac{a+2014+a-2014}{b+2015+b-2015}\)
\(=\dfrac{a+2014-\left(a-2014\right)}{b+2015-\left(b-2015\right)}=\dfrac{2a}{2b}=\dfrac{4028}{4030}=\dfrac{a}{b}=\dfrac{2014}{2015}\) (2)
Từ (2) : \(\dfrac{a}{b}=\dfrac{2014}{2015}\Rightarrow\dfrac{a}{2014}=\dfrac{b}{2015}\) ( đpcm )
Ta có: \(\dfrac{a+2014}{a-2014}=\dfrac{b+2015}{b-2015}\) ( \(a\ne\pm2014;b\ne\pm2015\))
\(\Rightarrow\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}=\dfrac{\left(a+2014\right)+\left(a-2014\right)}{\left(b+2015\right)+\left(b-2015\right)}=\dfrac{\left(a+2014\right)-\left(a-2014\right)}{\left(b+2015\right)-\left(b-2015\right)}=\dfrac{a+2014+a-2014}{b+2015+b-2015}=\dfrac{a+2014-a+2014}{b+2015-b+2015}=\dfrac{2a}{2b}=\dfrac{4018}{2030}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{2014}{2015}\)
\(\Rightarrow\dfrac{a}{2014}=\dfrac{b}{2015}\) (đpcm)