CMR nếu \(\dfrac{a+2014}{a-2014}=\dfrac{b+2015}{b-2015}\) thì \(\dfrac{a}{2014}=\dfrac{b}{2015}\)
vs \(a\ne\pm2014;b\ne\pm2015\)
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13 tháng 5 2022
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
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
AN
1
NQ
25 tháng 11 2017
Ta có : \(\frac{2014}{\sqrt{2015}}\)+ \(\frac{2015}{\sqrt{2014}}\) = \(\frac{2015-1}{\sqrt{2015}}\) + \(\frac{2014+1}{\sqrt{2014}}\)
= \(\sqrt{2015}\) + \(\sqrt{2014}\) + \(\frac{1}{\sqrt{2014}}\) - \(\frac{1}{\sqrt{2015}}\)
Vì \(\sqrt{2014}\) < \(\sqrt{2015}\) \(\Rightarrow \) \(\frac{1}{\sqrt{2014}}\)>\(\frac{1}{\sqrt{2015}}\) \(\Rightarrow \) \(\frac{1}{\sqrt{2014}}\)-\(\frac{1}{\sqrt{2015}}\) > 0
Nên \(\sqrt{2015}\) + \(\sqrt{2014}\) + \(\frac{1}{\sqrt{2014}}\) - \(\frac{1}{\sqrt{2015}}\) > \(\sqrt{2015}\) + \(\sqrt{2014}\)
Hay \(\frac{2014}{\sqrt{2015}}\)+ \(\frac{2015}{\sqrt{2014}}\) > \(\sqrt{2014} + \sqrt{2015}\)
14 tháng 4 2017
Ta có :
B = \(\dfrac{2015}{1}+\dfrac{2014}{2}+\dfrac{2013}{3}+...+\dfrac{2}{2014}+\dfrac{1}{2015}\) => B = \(\left(1+\dfrac{2014}{2}\right)+\left(1+\dfrac{2013}{3}\right)+...+\left(1+\dfrac{2}{2014}\right)+\left(1+\dfrac{1}{2015}\right)+1\) => B = \(\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2014}+\dfrac{2016}{2015}+\dfrac{2016}{2016}\) => B = \(2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\) Ta có :
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)}\)
=> \(\dfrac{A}{B}=\dfrac{1}{2016}\)
Vậy \(\dfrac{A}{B}=\dfrac{1}{2016}\)
12 tháng 9 2018
Với \(\forall a\in N\left(a\ne0\right)\cdot\),ta có:\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{\left(a^2+1\right)\left(a^2+2a+1\right)+a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{\left(a^2+1\right)^2+2a\left(a^2+1\right)+a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\dfrac{a^2+a+1}{a+1}+\dfrac{a}{a+1}=\dfrac{\left(a+1\right)^2}{a+1}=a+1\in Z\)(Vì a là số tự nhiên)
Thay a=2014 vào thì ta có: B=2014+1=2015 là số nguyên
\(\dfrac{a+2014}{a-2014}=\dfrac{b+2015}{b-2015}=\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}\) (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)