tra loi nhanh mk dang can

\(A=2008.\left(\frac{1}{2007}-\frac{1000}{1004}\right)-2009.\left(\frac{1}{2007}-2\right)\)

\(\Rightarrow A=2008.\frac{1}{2007}-2000-2009.\frac{1}{2007}+4018\)

\(\Rightarrow A=\frac{1}{2007}.\left(2008-2009\right)-2000+4018\)

\(\Rightarrow A=\frac{1}{2007}.\left(-1\right)-2000+4018\)

\(\Rightarrow A\approx2018\)

\(=\dfrac{2008}{2007}-2\cdot2009-\dfrac{2009}{2007}+2009\cdot2\)

=-1/2007

a: \(\left(2x-3\right)^{2012}+\left(y-\dfrac{2}{5}\right)^{2014}+\left|x+y-z\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y-\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{2}{5}\\z=\dfrac{19}{10}\end{matrix}\right.\)

b: 2015-|x-2015|=x

=>|x-2015|=2015-x

=>x-2015<=0

hay x<=2015

d: |x-999|+|1998-2x|=0

=>x-999=0

hay x=999

\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1003}\right)\)

\(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2016}\)

Đặt A=1-1/2+1/3-1/4+.......+1/2005-1/2006

=>A= (1+1/3+1/5+...+1/2005)-(1/2+1/4+1/6+.....+1/2006)

=>A=(1+1/2+1/3+...+1/2005)-2.(1/2+1/4+1/6+...+1/2006)

=>A=(1+1/2+1/3+....+1/2005)-(1+1/2+1/3+...+1/1003)

=>A=1/1004+1/1005+.....+1/2006

Vậy A=1/1004+1/1005+.....+1/2006 ( Điều phải chứng minh )

thỏa mãn là:bí kí xem sẽ

Do 2009 > 0 \(\Rightarrow f\left(2009\right)=2009+1=2010\)

DO -1004 < 0 \(\Rightarrow f\left(-1004\right)=1-2\cdot\left(-1004\right)=2009\)