2009<5*x-3<2013
xin loi minh viet nham de
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So sánh hai phân số :
\(\frac{2009.2009+2008}{2009.2009+2009}\)và \(\frac{2009.2009+2009}{2009.2009+2010}\)
Vì 20009 x 2009 + 2008 < 2009 x 2009 + 2009
=>A < 1
Ta có: \(B=\frac{2009x2009+2009}{2008x2009+2010}=\frac{2009x\left(2008+1\right)+2009}{2008x2009+2010}=\frac{2008x2009+2009+2009}{2008x2009+2010}\)
\(B=\frac{2008x2009+4018}{2008x2009+2010}=\frac{2008x2009+2010+2008}{2008x2009+2010}=\frac{2008x2009+2010}{2008x2009+2010}+\frac{2008}{2008x2009+2010}\)
\(B=1+\frac{2008}{2008x2009+2010}>1\)
Mà A < 1
=>A < B
Dễ quá, thực hiện qui tắc bỏ dấu ngoặc được:
\(2009+2009^2+....+2009^{2009}-1-2009-...-2009^{2008}\)
\(=-1+\left(2009-2009\right)+\left(2009^2-2009^2\right)+...+\left(2009^{2008}-2009^{2008}\right)+2009^{2008}\)
\(=2009^{2008}-1\)
\(=\left(2009-1\right)\left(2009^{2007}+2009^{2008}+...+2009+1\right)\)
\(=2008\left(2009^{2007}+2009^{2008}+...+2009+1\right)\) chia hết cho 2008
=> ĐPCM
Chứng Minh Rằng: (2009+20092+20093+20094+...+20092009)-(1+2009+20092+20093+...+20092008) chia hết cho 2008.
Đặt A=2009+20092+20093+20094+...+20092009, B=1+2009+20092+20093+20094+...+20092008
Ta có:
+)A=2009+20092+20093+20094+...+20092009
2009A= 20092+20093+20094+...+20092010
2009A-A=(20092+20093+20094+...+20092010)-(2009+20092+20093+20094+...+20092009)
2008A=20092010- 2009
=> A=(20092010- 2009)/2008
=> A chia hết cho 2008.
B=1+2009+20092+20093+20094+...+20092008
2009B=2009+20092+20093+20094+...+20092010
2009B-B=(2009+20092+20093+20094+...+20092010)-(1+2009+20092+20093+20094+...+20092009)
2008B=20092010-1
=>B=(20092010-1)/2008
=>B chia hết cho 2008
=> A-B chia hết cho 2008.
=> ĐPCM
\(2008-\dfrac{2009}{3}-\dfrac{2009}{6}-...-\dfrac{2009}{45}\\ =2008-2009\left(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{45}\right)\\ =2008-2009.2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right)\\ =2008-4018\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)\\ =2008-4018\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =2008-4018\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\\ =2008-4018.\dfrac{2}{5}=2008-1607,2\\ =400,8\)
a, \(\frac{1}{2009}+\frac{2}{2009}+...+\frac{2008}{2009}\\ \frac{\left(1+2008\right)\cdot2008\div2}{2009}=\frac{2017036}{2009}\)
Giải:
Ta có:
A=20092008+1/20092009+1
2009A=20092009+2009/20092009+1
2009A=20092009+1+2008/20092009+1
2009A=20092009+1/20092009+1 + 2008/20092009+1
2009A=1+2008/20092009+1
Tương tự:
B=20092009+1/20092010+1
2009B=1+2008/20092010+1
Vì 2008/20092009+1 > 2008/20092010+1 nên 2009A>2009B
⇒A>B
2009<5.x-3<2013
=> 2009+3<5.x-3+3<2013+3
=> 2012<5.x<2016
=>402<x<404
=> x=403
Sai đè
Vì nếu như vậy ko có c thỏa mãn