\(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2006.2007}=\frac{2006}{2007}\)

\(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+\frac{x}{3}-\frac{x}{4}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)

\(x-\frac{x}{2007}=\frac{2006}{2007}\)

\(\frac{2007x}{2007}-\frac{x}{2007}=\frac{2006}{2007}\)

\(2007x-x=2006\)

\(2006x=2006\)

\(x=1\)

theo mình x =1 không biết đúng không

theo suy luận của mình thì x sẽ bằng 1

Đặt \(A=\frac{2006}{1\cdot2}+\frac{2006}{2\cdot3}+...+\frac{2006}{2006\cdot2007}\)

\(2006A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2006\cdot2007}\)

\(2006A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\)

\(2006A=\frac{1}{1}-\frac{1}{2007}\)

\(2006A=\frac{2006}{2007}\)

\(A=\frac{2006}{2007}\div2006\)

\(A=\frac{1}{2007}\)

Vậy giá trị của biểu thức bằng 1/2007

* Không chắc nha * 

Sửa đề : \(A=\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)

\(2006A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\)

\(2006A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\)

\(2006A=1-\frac{1}{2007}\)

\(2006A=\frac{2006}{2007}\)

\(A=\frac{2006}{2007}:2006=\frac{2006}{2007}.\frac{1}{2006}=\frac{1}{2007}\)

Giúp mình với

\(N=\dfrac{2006}{1.2}+\dfrac{2006}{2.3}+...+\dfrac{2006}{2006.2007}\)

\(N.2006=\dfrac{2006}{1}-\dfrac{2006}{2}+\dfrac{2006}{2}-\dfrac{2006}{3}+...+\dfrac{2006}{2006}-\dfrac{2006}{2007}\)

\(N.2006=2006-\dfrac{2006}{2007}\)

\(N=2006-\dfrac{2006}{2007}:2006\)

\(N=2006-\dfrac{1}{2007}\)

Đúng là câu b sai, nhầm dấu đoạn đầu, phải là \(\frac{2006.2006-\left(2005.2006+2005\right)}{2006.\left(2007-2005\right)}\)

Phá ngoặc thì thành trừ nhưng cô của em bạn lại sót=> sai luôn cả tính chất bài toán.

P/s: Thử lại bằng casio là thấy rõ bạn đúng.

Đặt biểu thức là A ta có:

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

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

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

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

\(\Rightarrow A=\frac{2006}{2007}\)

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

Đặt N = \(\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}\)

\(\Rightarrow N=\frac{1}{2006}+.....+\frac{2004}{3}+\frac{2005}{2}+\frac{2006}{1}\)

\(\Rightarrow N=\left(\frac{1}{2006}+1\right)+.....+\left(\frac{2004}{3}+1\right)+\left(\frac{2005}{2}+1\right)+1\)( Có 2005 nhóm )

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

Đặt M = \(\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}\)

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

Thay N và M vào C , ta có :

\(C=\frac{N}{M}=\frac{2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)}{2007\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2007}\right)}=\frac{2006}{2007}\)

\(\Rightarrow C=\frac{2006}{2007}\)

Vậy : \(C=\frac{2006}{2007}\)