Ta có

\(\hept{\begin{cases}\sqrt{2008}+\sqrt{2005}< \sqrt{2015}+\sqrt{2009}\left(1\right)\\\sqrt{2010}+\sqrt{2007}< \sqrt{2015}+\sqrt{2009}\left(2\right)\end{cases}}\)

\(\Rightarrow\frac{1}{\sqrt{2008}+\sqrt{2005}}+\frac{1}{\sqrt{2010}+\sqrt{2007}}>\frac{2}{\sqrt{2015}+\sqrt{2009}}\)

\(\Leftrightarrow\frac{\sqrt{2008}-\sqrt{2005}}{3}+\frac{\sqrt{2010}-\sqrt{2007}}{3}>\frac{\sqrt{2015}-\sqrt{2009}}{3}\)

\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}\)

A=√2008+√2009+√2010A=2008+2009+2010 và B=√2005+√2007+√2015

k và kb với mình nha !!!

tính cụ thể từng cái ra

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

\(B=\sqrt{2008}-\sqrt{2007}=\frac{\left(\sqrt{2008}-\sqrt{2007}\right)\left(\sqrt{2008}+\sqrt{2007}\right)}{\left(\sqrt{2008}+\sqrt{2007}\right)}=\frac{1}{\sqrt{2008}+\sqrt{2007}}\)(2)

Từ 1  và 2 => \(\frac{1}{\sqrt{2007}+\sqrt{2006}}>\frac{1}{\sqrt{2008}+\sqrt{2007}}\)

hay \(\sqrt{2007}-\sqrt{2006}>\sqrt{2008}-\sqrt{2007}\)

P/s tham khảo nha

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

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

Áp dụng \(\sqrt{\frac{a+b}{2}}>\frac{\sqrt{a}+\sqrt{b}}{2}\) được \(\sqrt{\frac{2007+2005}{2}}>\frac{\sqrt{2005}+\sqrt{2007}}{2}\Rightarrow2\sqrt{2006}>\sqrt{2005}+\sqrt{2007}\)

\(A=\sqrt{2005}+\sqrt{2007}\Rightarrow A^2=\left(\sqrt{2005}+\sqrt{2007}\right)^2=2005+2007+2\sqrt{2005\cdot2007}=4012+2\sqrt{\left(2006-1\right)\left(2006+1\right)}=4012+2\sqrt{2006^2-1}\)

\(B=2\sqrt{2006}\Rightarrow B^2=\left(2\sqrt{2006}\right)^2=4\cdot2006=2\cdot2006+2\cdot2006=4012+2\sqrt{2006^2}\)

Ta thấy \(4012=4012\) và \(\sqrt{2006^2-1}< \sqrt{2006^2}\)
nên \(A^2< B^2\)\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
 

Có \(\sqrt{2005}+\sqrt{2007}=2005+2007+2\sqrt{2005\cdot2007}\)

                                            \(=2005+2007+2\sqrt{\left(2006-1\right)\left(2006+1\right)}\)

                                            \(=4012+2\sqrt{2006^2-1}\)

\(2\sqrt{2006}=2006+2006+2\cdot2006\)

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

Mà \(4012+2\sqrt{2006^2-1}< 4012+2\sqrt{2006^2}\)

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

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

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

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

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

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

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

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

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

ai biết thì giải giúp với 

bằng nhau

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

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

\(\sqrt{2006}=44,78839135\)