Theo mk thì hình như

=0 

Ko biết chắc !

\(\frac{1}{2007}.\left(\frac{1001}{2006}-2007\right)-\left(\frac{1}{2006}-2007\right).\frac{1001}{2007}\)

\(=\left(\frac{1001}{2007.2006}-\frac{2007}{2007}\right)-\left(\frac{1001}{2006.2007}-\frac{2007.1001}{2007}\right)\)

\(=\frac{1001}{2007.2006}-\frac{1001}{2006.2007}-1+1001\)

\(=-1+1001\)

\(=1000\)

\(ĐK:\left\{{}\begin{matrix}x-2008\ge0\\2008-x\ge0\\x-2007>0\end{matrix}\right.\Leftrightarrow x=2008\)

Vậy PT có nghiệm \(x=2008\)

Hình như thiếu mũ 2007 -.- Sửa luôn nhóe :)

Trước hết ta tính tổng sau, với các số tự nhiên a, n đều lớn hơn 1.

\(S_n=\dfrac{1}{a}+\dfrac{1}{a^2}+...+\dfrac{1}{a^n}\)

Ta có: \(\left(a-1\right)S_n=aS_n-S_n\)

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

Áp dụng BĐT ( 1 ) cho a = 2008 và mọi n = 2,3, ..., 2004 ta được:

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

Lại áp dụng BĐT ( 1 ) cho a = 2007 và n = 2007, ta được:

\(\dfrac{1}{2007}+\dfrac{1}{2007^2}+...+\dfrac{1}{2007^{2007}}< \dfrac{1}{2006}=A\left(3\right)\)

Từ ( 2 ) và ( 3 ) => B < A.

Thiệt ta là tui chép sách

số số hạng của A là :

( 2007 - 3 ) : 3 + 1 = 669 ( số )

tổng A là :

( 2007 + 3 ) . 669 : 2 = 672345

B = \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)

B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...+\left(\dfrac{1}{2006}+1\right)+1}\)

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

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

B = \(\dfrac{2006}{2007}\)

\(A=3+6+9+12+...+2007\)

\(A=\dfrac{\left(2007+3\right)\left(\dfrac{2007-3}{3}+1\right)}{2}\)

\(A=\dfrac{2010.669}{2}=672235\)

\(B=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)

\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{2006+\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...\left(\dfrac{1}{2006}+1\right)-2005}\)

\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2005}{3}+...+\dfrac{2007}{2006}}\)

\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{2007\left(\dfrac{1}{2007}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}\right)}=\dfrac{2006}{2007}\)

tik mik nha !!!

Nguyễn copy mình à

\(Q=\sqrt{1+2006^2+\left(\dfrac{2006}{2007}\right)^2}+\dfrac{2006}{2007}\)  

   =\(1+2006+\dfrac{2006}{2007}+\dfrac{2006}{2007}\)

   =\(2007+\dfrac{4012}{2007}\)

   =\(\dfrac{2007^2}{2007}+4012\)

   =\(\dfrac{4028049}{2007}+\dfrac{4012}{2007}\)

  =\(\dfrac{4032061}{2007}\)

\(Q=\sqrt{1+2006^2+\dfrac{2006^2}{2007^2}}+\dfrac{2006}{2007}\)

\(=1+2006+\dfrac{2006}{2007}+\dfrac{2006}{2007}\)

\(=\dfrac{4032061}{2007}\)

Áp dụng bđt AM-GM cho 2 số không âm ta có:

\(\dfrac{1}{\sqrt{1.2006}}>\dfrac{1}{\dfrac{1+2006}{2}}=\dfrac{2}{2007}\)

TT: \(\dfrac{1}{\sqrt{2.2005}}>\dfrac{2}{2007}\)

...

\(\dfrac{1}{\sqrt{2006.1}}>\dfrac{2}{2007}\)

Cộng vế với vế ta được:

\(S>\dfrac{2}{2007}.2006\)

ko đc tag tên có đc lm ko

Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)

\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)

\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)

\(=\dfrac{2006}{2007}\)

Đặt: \(L_2=\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)

\(L_2=1+\left(\dfrac{2006}{2}+1\right)+\left(\dfrac{2005}{3}+1\right)+...+\left(\dfrac{2}{2006}+1\right)+\left(\dfrac{1}{2007}+1\right)\)

\(L_2=\dfrac{2008}{2008}+\dfrac{2008}{2}+\dfrac{2008}{3}+...+\dfrac{2008}{2006}+\dfrac{2008}{2007}\)

\(L_2=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)

\(\dfrac{L_1}{L_2}=\dfrac{1}{2008}\)

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

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

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