A>b

Cách làm: Bạn tách |B ra rồi so sánh với từng ps ở A, sau đó Kết luận

1) Ta có bđt sau : \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m)

Áp dụng : \(\frac{\sqrt{2005}+\sqrt{2007}}{2}< \sqrt{\frac{2005+2007}{2}}\)

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

2) Xét : \(A-B=2\sqrt{2014}-\left(\sqrt{2013}+\sqrt{2015}\right)\)

Theo câu 1) , ta dễ dàng c/m được \(2\sqrt{2014}>\sqrt{2013}+\sqrt{2015}\)

Do đó A - B > 0 => A > B

2) Bình phương 2 vế ta có:

 \(A^2=2014-2013=1\)

\(B^2=2015-2014=1\)

=>A=B

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

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

\(ĐK:\left\{{}\begin{matrix}x-2008\ge0\\2008-x\ge0\\x-2007>0\end{matrix}\right.\Leftrightarrow x=2008\)

Vậy PT có nghiệm \(x=2008\)

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

Theo to:

A>B

\(A=\frac{2006+2007}{2006.2007}=\frac{2006}{2006.2007}+\frac{2007}{2006.2007}=\frac{1}{2007}+\frac{1}{2006}\)

\(B=\frac{2007+2008}{2007.2008}=\frac{2007}{2007.2008}+\frac{2008}{2007.2008}=\frac{1}{2008}+\frac{1}{2007}\)

Vì \(\frac{1}{2007}+\frac{1}{2006}>\frac{1}{2008}+\frac{1}{2007}\)

=> \(A>B\)

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: \(A< B\)