\(a,A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(=\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\)

\(=\frac{1-\sqrt{100}}{-1}=9\)

\(b,B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..+\frac{1}{\sqrt{99}}\)

\(=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{99}}>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)\(\Rightarrow B>2\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\right)\)

\(\Rightarrow B>2\left(\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\right)\)

\(\Rightarrow B>2\left(\frac{1-\sqrt{100}}{-1}\right)\)

\(\Rightarrow B>2.9=18\left(ĐPCM\right)\)

Ta có :

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

Ta có:

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

Do đó:

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)

\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)

(Fix luôn lại đề)

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

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

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

Bài 2:

Áp dụng bài 1 vào A được:

A\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

Ta có :

\(\hept{\begin{cases}\frac{1}{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\end{cases}}\forall n\in N\)

Suy ra : \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)

Đặt \(M=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2499}}+\frac{1}{\sqrt{2500}}\)

\(\Leftrightarrow\frac{1}{2}M=\frac{1}{2\sqrt{2500}}+\frac{1}{2\sqrt{2499}}+...+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{2}}+\frac{1}{2}\)

Áp dụng BĐT , ta có :

\(\frac{1}{2}M< \sqrt{2500}-\sqrt{2499}+\sqrt{2499}-\sqrt{2498}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}M< \sqrt{2500}-\sqrt{1}+\frac{1}{2}=50-\frac{1}{2}< 50\)

\(\Rightarrow M< 100\)

a, Chắc xét hàm số tổng quát!

Xét hàm số tổng quát:

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

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

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

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

Vì \(\dfrac{\sqrt{k}}{\sqrt{k+1}}< 1\Rightarrow1+\dfrac{\sqrt{k}}{\sqrt{k+1}}< 2\)

Do đó \(\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)< 2.\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)

\(\Rightarrow\dfrac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\) (1)

Áp dụng điểu (1) ta được:

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

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

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

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

\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+....+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)

Với mọi giá trị của \(n>0\) ta luôn có: \(\sqrt{n+1}>0\)

Do đó \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\) (đpcm)

Mấy bài này đã có người làm rồi nhé bạn vào câu hỏi tương tự mà xem.