\(\dfrac{2006}{1.2}+\dfrac{2006}{2.3}+...+\dfrac{2006}{2006.2007}\)
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6 tháng 5 2020
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 *
6 tháng 5 2020
Sửa đề : \(A=\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)
\(2006A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\)
\(2006A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\)
\(2006A=1-\frac{1}{2007}\)
\(2006A=\frac{2006}{2007}\)
\(A=\frac{2006}{2007}:2006=\frac{2006}{2007}.\frac{1}{2006}=\frac{1}{2007}\)
HT
3
18 tháng 3 2017
\(\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)
\(=2006.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\right)\)
\(=2006.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\right)\)
\(=2006.\left(1-\frac{1}{2007}\right)\)
\(=2006.\frac{2006}{2007}\)
\(=\frac{2006^2}{2007}\)
HL
18 tháng 3 2017
\(=\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)
\(=2006 \left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\right)\)
\(=2006.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\right)\)
\(=2006.\left(1-\frac{1}{2007}\right)\)
\(=2006.\frac{2006}{2007}=\frac{4024036}{2007}\)
12 tháng 5 2021
Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)
\(=\dfrac{2006}{2007}\)
21 tháng 6 2022
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
4 tháng 8 2017
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 = \(\dfrac{2006}{2007}\)
4 tháng 8 2017
2006/1 là 2006, tách 1 của 2006 ra 2005 phân số còn lại 1
TN
4
7 tháng 4 2016
\(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2006.2007}=\frac{2006}{2007}\)
\(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+\frac{x}{3}-\frac{x}{4}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)
\(x-\frac{x}{2007}=\frac{2006}{2007}\)
\(\frac{2007x}{2007}-\frac{x}{2007}=\frac{2006}{2007}\)
\(2007x-x=2006\)
\(2006x=2006\)
\(x=1\)
14 tháng 2 2018
Áp dụng Bất đẳng thức :
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
Ta có :
\(\dfrac{2006^{2006}+1}{2006^{2007}+1}< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}=\dfrac{2006^{2006}+2006}{2006^{2007}+2006}=\dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}=\dfrac{2006^{2005}+1}{2006^{2006}+1}\)
\(\Leftrightarrow\dfrac{2006^{2006}+1}{2006^{2007}+1}< \dfrac{2006^{2005}+1}{2006^{2006}+1}\)
Giúp mình với
\(N=\dfrac{2006}{1.2}+\dfrac{2006}{2.3}+...+\dfrac{2006}{2006.2007}\)
\(N.2006=\dfrac{2006}{1}-\dfrac{2006}{2}+\dfrac{2006}{2}-\dfrac{2006}{3}+...+\dfrac{2006}{2006}-\dfrac{2006}{2007}\)
\(N.2006=2006-\dfrac{2006}{2007}\)
\(N=2006-\dfrac{2006}{2007}:2006\)
\(N=2006-\dfrac{1}{2007}\)