\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{99}+\sqrt{100}\)

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

\(=10-1\)

\(=9\)

Vì 9 chia hết cho 1; 3; 9 nên ko thể là số nguyên tố mà là hợp số.

=> ĐPCM

Bạn ơi xem kĩ lại đề bài đi

Ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(...........\)

\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

Cộng theo vế ta có:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\) (Đpcm)

Ta có : \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}.100=10\)

=>ĐPCM

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

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

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

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

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

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

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

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

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

Ta có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(VT>\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}+\frac{1}{\sqrt{100}}\)

\(=\frac{1}{10}+\frac{1}{10}\) có 100 số hạng 

\(=\frac{100}{10}=10\)

Dòng 6 cuối cùng mình làm cũng không được chắc chắn lắm đâu òng 6 đấy bạn ngoặc ở dưới 1/10 +1/10 nhé

Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

.......

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

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=10\) (đpcm)

Ap dung \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Ta co \(P< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2008}}\right)\)  

=> \(P< 2\left(1-\frac{1}{\sqrt{2008}}\right)< 2.1=2\)

Suy ra P khong phai so nguyen to