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

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

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

Bài 2:

\(D=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{120\sqrt{121}+121\sqrt{120}}\)

Với mọi \(n\inℕ^∗\)ta có:

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

\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}]^2-\left(n\sqrt{n+1}\right)^2}\)

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

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

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

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

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

\(\Rightarrow D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}\)

\(=1-\frac{1}{\sqrt{121}}=\frac{10}{11}\)

Bài 1: chắc lại phải "liên hợp" gì đó rồi:V

\(\sqrt{2009}-\sqrt{2008}=\frac{1}{\sqrt{2009}+\sqrt{2008}}\)

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

Đó \(\sqrt{2009}+\sqrt{2008}>\sqrt{2007}+\sqrt{2006}\)

Nên \(\sqrt{2009}-\sqrt{2008}< \sqrt{2007}-\sqrt{2006}\)

Tổng quát ta có bài toán sau, với So sánh \(\sqrt{n}-\sqrt{n-1}\text{ và }\sqrt{n-2}-\sqrt{n-3}\)

Với \(n\ge3\). Lời giải xin mời các bạn:)

Câu a) 

Có: \(A=\sqrt{2009}-\sqrt{2008}\Leftrightarrow A^2=1-2\sqrt{2009\cdot2008}\)

\(B=\sqrt{2007}-\sqrt{2006}\Rightarrow B^2=1-2\sqrt{2007\cdot2006}\)

Đương nhiên: \(2\sqrt{2009\cdot2008}>2\sqrt{2006\cdot2007}\)

Suy ra: \(A< B\)

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

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

\(b,\) Ta có:

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

Thay:

\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)

\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)

\(...\)

\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)

Tiếp phần b ( do máy lag) :3

Cộng 2 vế với nhau, ta có:

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\\ =1-\dfrac{1}{\sqrt{2007}}\)

a) A=\(\dfrac{1}{\sqrt{3}+\sqrt{5}}\)+\(\dfrac{1}{\sqrt{5}+\sqrt{7}}\)+\(\dfrac{1}{\sqrt{7}+\sqrt{9}}\)+...+\(\dfrac{1}{\sqrt{97}+\sqrt{99}}\)

=\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)+\(\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)+\(\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\left(\sqrt{9}-\sqrt{7}\right)}\)+...+\(\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}\)

=\(\dfrac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}}{2}\)

=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)

vậy A=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)