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

Đặt: n=2014

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

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

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

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

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

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

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

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

CHTT bạn.
bài này cí cttq nhé.xem sbt

a,a=b+1

suy ra a-b=1 suy ra(\(\sqrt{a}+\sqrt{b}\))(\(\sqrt{a}-\sqrt{b}\))=1

suy ra \(\sqrt{a}-\sqrt{b}\)=\(\frac{1}{\sqrt{a}+\sqrt{b}}\)(1)

vì a=b+1 suy ra a>b suy ra \(\sqrt{a}>\sqrt{b}\)suy ra \(\sqrt{a}+\sqrt{b}>2\sqrt{b}\)

suy ra \(\frac{1}{\sqrt{a}+\sqrt{b}}< \frac{1}{2\sqrt{b}}\)(2)

từ (1) ,(2) suy ra\(\sqrt{a}-\sqrt{b}< \frac{1}{2\sqrt{b}}\)suy ra \(2\left(\sqrt{a}-\sqrt{b}\right)< \frac{1}{\sqrt{b}}\)(*)

ta lại có b+1=c+2 suy ra b-c =1 suy ra\(\left(\sqrt{b}-\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)=1\)

suy ra \(\sqrt{b}-\sqrt{c}=\frac{1}{\sqrt{b}+\sqrt{c}}\)(3)

vì b>c suy ra \(\sqrt{b}>\sqrt{c}\) suy ra \(\sqrt{b}+\sqrt{c}>2\sqrt{c}\)

suy ra \(\frac{1}{\sqrt{b}+\sqrt{c}}< \frac{1}{2\sqrt{c}}\)(4)

Từ (3),(4) suy ra \(\sqrt{b}-\sqrt{c}< \frac{1}{2\sqrt{c}}\) suy ra\(2\left(\sqrt{b}+\sqrt{c}\right)< \frac{1}{\sqrt{c}}\)(**)

từ (*),(**) suy ra đccm

a) Có \(A=\frac{\sqrt{x}+2}{\sqrt{x}-2}=\frac{\sqrt{x}-2+4}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}=1+\frac{4}{\sqrt{x}-2}\)

Để A đạt giá trị nguyên thì: \(\sqrt{x}-2\in U\left(4\right)\)

TH1: \(\sqrt{x}-2=1\Rightarrow x=9\)

TH2: \(\sqrt{x}-2=-1\Rightarrow x=1\)

TH3: \(\sqrt{x}-2=2\Rightarrow x=16\)

TH4: \(\sqrt{x}-2=-2\Rightarrow x=0\)

TH5: \(\sqrt{x}-2=4\Rightarrow x=36\)

TH6: \(\sqrt{x}-2=-4\Rightarrow\) k tồn tại x

Vậy:...

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

ki+e

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