Easy

Ta có:

\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)

Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)

Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)

\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)

Easy

Ta có:

\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)

Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)

Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)

\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)

Ta có : \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)

             \(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)

Mà : \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}-\sqrt{2006}}\)

Nến : \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)

\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)

\(\left(\sqrt{2005}+\sqrt{2007}\right)^2=4012+2\sqrt{2005.2007}\)

\(=4012+2\sqrt{\left(2016-1\right)\left(2016+1\right)}=4012+2\sqrt{2016^2-1}\)

\(\left(2\sqrt{2006}\right)^2=4012+4012=4012+2\sqrt{2016^2}\)

=>\(\left(\sqrt{2015}+\sqrt{2017}\right)^2< \left(2\sqrt{2016}\right)^2\Rightarrow\sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}\)

Ta có: \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)

Mà: \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2006}}\)

Nên: \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)

=>\(\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)

k đi mình làm cho

\(\sqrt{2004}-\sqrt{2003}=\dfrac{1}{\sqrt{2004}+\sqrt{2003}}\)

\(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

Mà \(\sqrt{2004}+\sqrt{2003}< \sqrt{2006}< \sqrt{2005}\)

\(\Rightarrow\dfrac{1}{\sqrt{2004}+\sqrt{2003}}>\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\Rightarrow\sqrt{2004}-\sqrt{2003}>\sqrt{2006}-\sqrt{2005}\)

a/ giả sử \(\sqrt{7}-\sqrt{2}< 1\)

\(\Leftrightarrow\sqrt{7}< 1+\sqrt{2}\)

\(\Leftrightarrow 7< 1+2\sqrt{2}+2\)

\(\Leftrightarrow4< 2\sqrt{2}\Leftrightarrow16< 8\left(sai\right)\)

vậy \(\sqrt{7}-\sqrt{2}>1\)

câu b, c bạn làm tương tụ nhé , giả sử một đẳng thức tạm, sau đó bình phương lên rồi làm theo như trên là được nha 

Bài này cũng dễ

a, \(\sqrt{7}-\sqrt{2}\) lớn hơn \(1\) . Vì

\(\sqrt{7}-\sqrt{2}=1,231537749\)

\(1=1\)

b, \(\sqrt{8}+\sqrt{5}\) bé hơn \(\sqrt{7}+\sqrt{6}\) . Vì

\(\sqrt{8}+\sqrt{5}=5,064495102\) 

\(\sqrt{7}+\sqrt{6}=5,095241054\)

c, \(\sqrt{2005}+\sqrt{2007}\) lớn hơn \(\sqrt{2006}\) . Vì

\(\sqrt{2005}+\sqrt{2007}=89,57677992\)

\(\sqrt{2006}=44,78839135\) 

\(\sqrt{2005+2006}^2=2005+2006=4011\)

\(\left(\sqrt{2005}+\sqrt{2006}\right)^2=2005+2\sqrt{2005}.\sqrt{2006}+2006=4011+2\sqrt{2005}.\sqrt{2006}\)

Vì \(2\sqrt{2005}.\sqrt{2006}>0\) nên =>\(4011

ai biết thì giải giúp với 

bằng nhau

\(\sqrt{2007}-\sqrt{2006}=\frac{\sqrt{2007}-\sqrt{2006}}{2007-2006}=\frac{\sqrt{2007}-\sqrt{2006}}{\left(\sqrt{2007}-\sqrt{2006}\right)\left(\sqrt{2007}+\sqrt{2006}\right)}\)

\(=\frac{1}{\sqrt{2007}+\sqrt{2006}}< \frac{1}{\sqrt{2006}+\sqrt{2006}}=\frac{1}{2\sqrt{2006}}\)

Vậy \(\sqrt{2007}-\sqrt{2006}< \frac{1}{2\sqrt{2006}}\)

Bạn áp dùng biểu thức liên hợp là được

Ta có :

\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)(1)

\(\frac{1}{2\sqrt{2006}}=\frac{1}{\sqrt{2006}+\sqrt{2006}}\)(2)

Từ (1)(2)=>\(\frac{1}{\sqrt{2007}+\sqrt{2006}}< \frac{1}{\sqrt{2006}+\sqrt{2006}}\)

\(\Rightarrow\sqrt{2007}-\sqrt{2006}>\frac{1}{2\sqrt{2006}}\)