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14 tháng 7 2015

2015=2014+1 => 2015^2=2014^2 +2.2014+1

=>2014^2 + 1=2015^2 -2.2014=2015^2 -2. 2014/2015.2015

thế vào =>b= 2015-2014/2015+2014/2015=2015

15 tháng 10 2016

Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014

15 tháng 10 2016

ki+e

n ejmfjnhcy

Ta có \(2015^2=\left(2014+1\right)^2=2014^2+2.2014+1\) 

=> \(2014^2+1=2015^2-2.2014\) 

=> \(B=\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\) 

\(\sqrt{2015^2-2.2014+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\) 

\(\sqrt{\left(2015-\frac{2014}{2015}\right)^2}+\frac{2014}{2015}\) = \(2015-\frac{2014}{2015}+\frac{2014}{2015}\) 

\(2015\) là số nguyên

=> đpcm

29 tháng 9 2018

Đặt: n=2014

Ta có: \(1+n^2+\left(\frac{n}{n+1}\right)^2=\frac{\left(n+1\right)^2+n^2\left(n+1\right)^2+n^2}{\left(n+1\right)^2}\)

\(=\frac{\left(n+1\right)^2+n^2\left(n^2+2n+2\right)}{\left(n+1\right)^2}=\frac{\left(n+1\right)^2+2n^2\left(n+1\right)+n^4}{\left(n+1\right)^2}\)

\(=\frac{\left(n^2+n+1\right)^2}{\left(n+1\right)^2}=\left(\frac{n\left(n+1\right)+1}{n+1}\right)^2=\left(n+\frac{1}{n+1}\right)^2\)

\(\Rightarrow\sqrt{1+n^2+\left(\frac{n}{n+1}\right)^2}=n+\frac{1}{n+1}\)

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

27 tháng 7 2017

Ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

b, Ta có \(2015^2=\left(2014+1\right)^2=2014^2+2.2014+1\) 

=> \(2014^2+1=2015^2-2.2014\) 

=> \(B=\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\) 

\(\sqrt{2015^2-2.2014+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\) 

\(\sqrt{\left(2015-\frac{2014}{2015}\right)^2}+\frac{2014}{2015}\) = \(2015-\frac{2014}{2015}+\frac{2014}{2015}=2015\) 

=> đpcm

31 tháng 10 2017

ta có: \(A=\sqrt{1+2.2014+2014^2-2.2014+\frac{2014^2}{2015^2}}+\frac{2014}{2015}.\)

\(A=\sqrt{2015^2-2.2015.\frac{2014}{2015}+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\)

\(A=\sqrt{\left(2015-\frac{2014}{2015}\right)^2}+\frac{2014}{2015}\)

\(A=2015-\frac{2014}{2015}+\frac{2014}{2015}=2015\)

Vậy A=2015

16 tháng 1 2016

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

\(=2014\left(1+\frac{1}{2014}+\frac{1}{2015}\right)-\frac{2014}{2015}=2015\)

16 tháng 1 2016

\(B=\sqrt{2014^2\left(1+\frac{1}{2014}-\frac{1}{2015}\right)^2}+\frac{2014}{2015}=2015\)

4 tháng 10 2020

Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành: 

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)

\(\Rightarrow x=2018;y=2019;z=2020\)

4 tháng 10 2020

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)

\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)

\(x=2018,y=2019,z=2020\)

1 tháng 9 2019

Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)

\(\sqrt{y-2014}=b\left(b>0\right)\)

\(\sqrt{z-2015}=c\left(c>0\right)\)

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)

<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)

<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)

<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).

\(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)

\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)

\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)

=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)

Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)

1 tháng 9 2019

ko bt