1)\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)

\(\dfrac{A}{B}=\dfrac{1}{2009}\)

2) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)

\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)

Ta có :

\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=2009\)

\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)

\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)

\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)

\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)

\(\dfrac{A}{B}=\dfrac{1}{2009}\)

\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

=>\(\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

=>x-2010=0

=>x=2010

(x - 1)/2009 + (x - 2)/2008 = (x - 3)/2007 + (x - 4)/2006

(x - 1)/2009 - 1 + (x - 2)/2008 - 1 = (x - 3)/2007 - 1 + (x - 4)/2006 - 1

(x - 2010)/2009 + (x - 2010)/2008 = (x - 2010)/2007 + (x - 2010)/2006

(x - 2010)/2009 + (x - 2010)/2008 - (x - 2010)/2007 - (x - 2010)/2006 = 0

(x - 2010).(1/2009 + 1/2008 - 1/2007 - 1/2006) = 0

x - 2010 = 0

x = 2010

\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\\\Rightarrow\dfrac{x-1}{2009}+\dfrac{x-2}{2008}-\dfrac{x-3}{2007}-\dfrac{x-4}{2006}=0\\\Rightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)-\left(\dfrac{x-3}{2007}-1\right)-\left(\dfrac{x-4}{2006}-1\right)=0 \\ \Rightarrow\dfrac{x-1-2009}{2009}+\dfrac{x-2-2008}{2008}-\dfrac{x-3-2007}{2007}-\dfrac{x-4-2006}{2006}=0\\ \Rightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\\\Rightarrow \left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\\ \)
Mà \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
\(\Rightarrow x-2010=0\\ \Rightarrow x=2010\)

Vậy \(x=2010\)

Lời giải:

a) 

PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)

\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)

\(\Leftrightarrow (x+2)^3-4(x+2)(x^2-2x+4)=0\)

\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)

\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)

\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)

b) Bạn kiểm tra lại xem có sai đề không?

\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)

=>x+2011=0

hay x=-2011

Đặt D1 = \(\dfrac{2010}{1}\) + \(\dfrac{2009}{2}\) + \(\dfrac{2008}{3}\) + ... + \(\dfrac{1}{2010}\)

= 1 + ( 1+ \(\dfrac{2009}{2}\)) + ( 1+ \(\dfrac{2008}{3}\)) + ... + (1+\(\dfrac{1}{2010}\))

= \(\dfrac{2011}{2}\) + \(\dfrac{2011}{3}\)+ ... + \(\dfrac{2011}{2010}\) + \(\dfrac{2011}{2011}\)

= 2011. ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\))

Đặt D2 = \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\)

=> D = 2011

cho mk 1 tick nha

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)