Question

[nguyen manh duc]

a)tính tổng S=\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+..+\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\)

b)Áp dụng, tìm phần nguyên của A=\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+...+\dfrac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\) với n lẻ

Answer 1

[DƯƠNG PHAN KHÁNH DƯƠNG]

Câu a : Ta có :

\(\dfrac{1}{1+\sqrt{2}}=\dfrac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}=\dfrac{1-\sqrt{2}}{1-2}=\dfrac{1-\sqrt{2}}{-1}=-1+\sqrt{2}\)

\(\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\dfrac{\sqrt{2}-\sqrt{3}}{2-3}=\dfrac{\sqrt{2}-\sqrt{3}}{-1}=-\sqrt{2}+\sqrt{3}\)

.....................

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

Thay vào ta được :

\(S=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...........-\sqrt{n^2-1}+\sqrt{n^2}\)

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

Answer 2

[Akai Haruma]

Câu b:

Đặt biểu thức đã cho là $A$

Ta có:

\(A>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\right)\)

\(\Leftrightarrow A> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\right)\)

\(\Leftrightarrow A> \frac{1}{2}(n-1)\) (áp dụng cách tính toán phần a)

Lại có:

\(A< \frac{1}{2}\left(\frac{1}{0+\sqrt{1}}+\frac{1}{1+\sqrt{2}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}\right)+....+\frac{1}{2}\left(\frac{1}{\sqrt{n^2-3}+\sqrt{n^2-2}}+\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\right)\)

\(\Leftrightarrow A< \frac{1}{2}\left(\frac{1}{0+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\right)\)

\(\Leftrightarrow A< \frac{\sqrt{n^2-1}}{2}\) (áp dụng cách tính toán của phần a)

Vậy \(\frac{\sqrt{n^2-1}}{2}> A> \frac{n-1}{2}\) hay \(\sqrt{t(t+1)}> A> t\) (đặt \(n=2t+1\) )

Mà \(\sqrt{t(t+1)}< \sqrt{(t+1)(t+1)}=t+1\)

Do đó: \(t+1> A> t\)

\(\Rightarrow \lfloor{A}\rfloor=t=\frac{n}{2}\)