c. Ta có: C+E=\(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}=\sqrt{\left(\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{41}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{49}{2}}-\sqrt{\dfrac{41}{2}}\right)^2}=\dfrac{7}{\sqrt{2}}+\dfrac{\sqrt{41}}{\sqrt{2}}+\dfrac{7}{\sqrt{2}}-\dfrac{\sqrt{41}}{\sqrt{2}}=\dfrac{2.7}{\sqrt{2}}=7\sqrt{2}\)

=> đpcm.

Bài 1: 

Ta có: \(a+b\ge2\sqrt{ab}\)

\(b+c\ge2\sqrt{bc}\)

\(a+c\ge2\sqrt{ac}\)

Do đó: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)

hay \(a+b+c\ge\sqrt{ab}+\sqrt{cb}+\sqrt{ac}\)

Lấy vế trái trừ vế phải ta có:

\(\frac{2010}{\sqrt{2009}}+\frac{2009}{\sqrt{2010}}-\sqrt{2009}-\sqrt{2010}=\)\(\frac{2010}{\sqrt{2009}}+\frac{2009}{\sqrt{2010}}-\frac{2009}{\sqrt{2009}}-\frac{2010}{\sqrt{2010}}\)=\(\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2010}}\) (1)

2009<2010 lên biểu thức (1) >0

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+........+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=\frac{1}{\sqrt{1}\sqrt{2}\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{\sqrt{2}\sqrt{3}\left(\sqrt{2}+\sqrt{3}\right)}+........+\frac{1}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2009}+\sqrt{2010}\right)}\)

\(=\frac{\left(\sqrt{2010}-\sqrt{2009}\right)\left(\sqrt{2010}+\sqrt{2009}\right)}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2010}+\sqrt{2009}\right)}+.......+\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}=1-\frac{1}{\sqrt{2010}}=1-\frac{\sqrt{2010}}{2010}\)

Sorry thiếu với \(\forall m\inℝ\)

với cả  : P(x) = ax² + bx +c , a khác 0

Ta có các số trong dãy đều có dạng 1/[ (n + 1)√n ]
Ta có: 1/[ (n + 1)√n ] = (√n)/[ (n + 1)√n.√n ] = (√n)/[ (n + 1)n ] = (√n).1/[ (n + 1)n ]
Do 1/[ (n + 1)n ] = 1/n - 1/(n + 1) (mình nghĩ bạn biết cái này)
=> (√n).1/[ (n + 1)n ] = (√n).[ 1/n - 1/(n + 1) ]
Ta có 1/n - 1/(n + 1) = (1/√n)² - [ 1/√(n + 1) ]²
= [ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ]
=> 1/n - 1/(n + 1) = [ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ]
=> (√n).[ 1/n - 1/(n + 1) ] = (√n).[ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ]
Nhân √n với [ 1/√n + 1/√(n + 1) ] ta được
(√n).[ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ] = [ 1 + (√n)/√(n + 1) ].[ 1/√n - 1/√(n + 1) ]
=> 1/[ (n + 1)√n ] = [ 1 + (√n)/√(n + 1) ].[ 1/√n - 1/√(n + 1) ] (1)
Do (√n)/√(n + 1) < √(n + 1)/√(n + 1)
=> (√n)/√(n + 1) < 1
=> 1 + (√n)/√(n + 1) < 1 + 1
=> 1 + (√n)/√(n + 1) < 2
=> [ 1 + (√n)/√(n + 1) ].[ 1/√n - 1/√(n + 1) ] < 2.[ 1/√n - 1/√(n + 1) ] (2)
Từ (1) và (2) => 1/[ (n + 1)√n ] < 2.[ 1/√n - 1/√(n + 1) ]
Áp dụng ta được
1/2 < 2( 1 - 1/√2)
1/3√2 < 2(1/√2 - 1/√3)
....
1/(n+1)√n < 2(1/√n - 1/√(n + 1) )
=> 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2( 1 - 1/√2) + 2(1/√2 - 1/√3) + ... + 2(1/√n - 1/√(n + 1) )
=> 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2( 1 - 1/√2 + 1/√2 - 1/√3 + ... + 1/√n - 1/√(n + 1) )
=> 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2(1 - 1/√(n + 1) ) (3)
Do 1√(n + 1) > 0
=> -1√(n + 1) < 0
=> 1 -1√(n + 1) < 1
=> 2(1 - 1/√(n + 1) ) < 2 (4)
Từ (3) và (4) => 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2

\(\dfrac{1}{\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n\left(n+1\right)}}.\dfrac{1}{\sqrt{n+1}}\) . Do \(\sqrt{n+1}>\dfrac{\sqrt{n}+\sqrt{n+1}}{2}\)

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

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

Áp dụng vào bài toán:

\(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{2009\sqrt{2008}}< 2\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2008}}-\dfrac{1}{\sqrt{2009}}\right)\)

\(\Rightarrow VT< 2\left(1-\dfrac{1}{\sqrt{2009}}\right)< 2-\dfrac{2}{\sqrt{2009}}< 2\) (đpcm)

Xét với n là số tự nhiên không nhỏ hơn 1 , ta có 

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

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

Áp dụng điều trên : 

\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2010\sqrt{2009}}< \)

\(< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2010}}\right)=2\left(1-\frac{1}{\sqrt{2010}}\right)< \)

\(< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)