Ghi nhầm 

\(\sqrt{3}+1<\sqrt{4}+1=3\)

Vậy 3 > \(\sqrt{3}+1\)

1/√1 > 1/10
1/√2 > 1/10
1/√3 > 1/10
....................
1/√99 > 1/10
1/√100 = 1/10
Cộng từng vế ta có:
1/√1 + 1/√2 + 1/√3 + ... + 1/√100 >100.1/0 = 10 (Đpcm)

\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10=\sqrt{100}>\sqrt{99}\)

Lời giải:
Đặt \(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{2004}}\)

Xét số hạng tổng quát: \(\frac{1}{\sqrt{n}}\) ta có:

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

Do đó:

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

\(\frac{1}{\sqrt{2}}> 2(\sqrt{3}-\sqrt{2})\)

\(\frac{1}{\sqrt{3}}> 2(\sqrt{4}-\sqrt{3})\)

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

\(\frac{1}{\sqrt{2004}}> 2(\sqrt{2005}-\sqrt{2004})\)

Cộng theo vế:
$A>2(\sqrt{2005}-1)>86$

Vậy..........

Ta có : \(\sqrt{17}+\sqrt{26}+1<\sqrt{16}+\sqrt{25}+1\)

=>\(\sqrt{17}+\sqrt{26}+1<10\)

Mà \(\sqrt{99}<10\) =>\(\sqrt{99}<\sqrt{17}+\sqrt{26}+1\)

Huỳnh Đức Lê viết sai dấu rồi

Ta có: \(A=\frac{\sqrt{3-\sqrt5}\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}\)

\(=\frac{\sqrt{6-2\sqrt5}\left(3+\sqrt5\right)}{\sqrt2\cdot\sqrt2\left(\sqrt5+1\right)}=\frac{\left(\sqrt5-1\right)\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}\)

\(=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}=1\)

Ta có: \(3-\sqrt5-1=2-\sqrt5<0\)

=>\(3-\sqrt5<1\)

=>\(\sqrt{3-\sqrt5}<1\)

=>B<1

Do đó: B<A

B=\(\sqrt{17}+\sqrt{5}+1\)>\(\sqrt{16}+\sqrt{4}+1\)=4+2+1=7=\(\sqrt{49}\)>\(\sqrt{45}\)

Vậy B>C