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

\(>4n^2+4n=4n\left(n+1\right)\)

\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)

\(\Rightarrow\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{1}{2}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\) \(=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)

\(< \frac{1}{2}\cdot\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}\right)\)

\(< \frac{1}{2}\)

Lời giải:

a) Ta thấy: \(a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\geq 0, \forall a,b>0\)

\(\Rightarrow a+b\geq 2\sqrt{ab}>0\Rightarrow \frac{1}{a+b}\le \frac{1}{2\sqrt{ab}}\).

Vì $a> b$ nên dấu bằng không xảy ra . Tức \(\frac{1}{a+b}< \frac{1}{2\sqrt{ab}}\)

Ta có đpcm

b)

Áp dụng kết quả phần a:

\(\frac{1}{3}=\frac{1}{1+2}< \frac{1}{2\sqrt{2.1}}\)

\(\frac{1}{5}=\frac{1}{3+2}< \frac{1}{2\sqrt{2.3}}\)

\(\frac{1}{7}=\frac{1}{4+3}< \frac{1}{2\sqrt{4.3}}\)

.....

\(\frac{1}{4021}=\frac{1}{2011+2010}< \frac{1}{2\sqrt{2011.2010}}\)

Do đó:

\(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)

\(< \frac{\sqrt{2}-\sqrt{1}}{2\sqrt{2.1}}+\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3.2}}+\frac{\sqrt{4}-\sqrt{3}}{2\sqrt{4.3}}+....+\frac{\sqrt{2011}-\sqrt{2010}}{2\sqrt{2011.2010}}\)

\(=\frac{1}{2}-\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}-\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{2010}}-\frac{1}{2\sqrt{2011}}\)

\(=\frac{1}{2}-\frac{1}{2\sqrt{2011}}< \frac{1}{2}\) (đpcm)

https://hoc24.vn/hoi-dap/question/817465.html

Bn tham khảo ở đây nha!

Mk lm r, k muốn lm lại

\(\Leftrightarrow\)A=\(\left|x-2010\right|+\left|x-2011\right|\)=\(\left|x-2010\right|+\left|2011-x\right|\)\(\ge\)\(\left|x-2010+2011-x\right|\)=1

Dấu ''='' xảy ra khi và chỉ khi \(\hept{\begin{cases}x-2010\ge0\\2011-x\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge2010\\x\le2011\end{cases}}\)\(\Leftrightarrow\)\(2010\le x\le2011\)

Vậy Min A =1 \(\Leftrightarrow2010\le x\le2011\)

chịu !!!