\(\frac{2008\times2009+1000}{2009\times2009-1009}=\frac{2008\times2009+1000}{2008\times2009+2009-1009}=\frac{2008\times2009+1000}{2008\times2009+1000}=1\)

2009² - 2018.2009 + 1009²

= 2009² - 2.2009.1009 + 1009²

= (2009 - 1009)² = 1000² = 1 000 000

Vậy giá trị của bt 2009² - 2018.2009 + 1009² có 6 chữ số 0

theo mình là 6 nha

\(\dfrac{2007x2010-1007}{2008x2009-1009}=\dfrac{2007x2009+2007-1007}{2007x2009+2009-1009}=\dfrac{2007x2009+1000}{2007x2009+1000}=1\)

2007x2010−1007​=2007x2009+2009−10092007x2009+2007−1007​=2007x2009+10002007x2009+1000​=1

đề cậu ơi

đề đó ạ chị giải giúp e\

 đề là tính nhanh ạ

= \(2009^2+2\cdot2009\cdot1009+1009^2\)

=\(\left(2009+1009\right)^2\)

=\(3018^2=9108324\)

2009² - 2018.2009 + 1009²

= 2009² - 2.2009.1009 + 1009²

= (2009 - 1009)²

= 1000² = 1 000 000

=(2009-1009)2
=1000²
=1000000

= 2009 * ( 2011 - 1 ) - 1000 / 2011 * 2009 - 1009                                         

= 2009 * 2011 - 2009 -1000 / 2011 * 2009 - 1009                                                                                              

= 2009 * 2011 - 1009 / 2011 * 2009 - 1009                                                                                                      

= 1 

\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)

\(\Leftrightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)

\(\left(3x-7\right)^{2007}.\left[\left(3x-7\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\\left(3x-7\right)=\pm1\end{cases}}}\)

=> \(x=\frac{7}{3},x=2,x=\frac{8}{3}\)

Vậy ...

2/\(\frac{5^{102}.9^{1009}}{3^{2018}.25^{50}}=\frac{5^{100+2}.3^{2.1009}}{3^{2018}.5^{2.50}}=\frac{5^{100}.5^2.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)

a, \(\frac{1}{2009}+\frac{2}{2009}+...+\frac{2008}{2009}\\ \frac{\left(1+2008\right)\cdot2008\div2}{2009}=\frac{2017036}{2009}\)