cho biểu thức :
A=\(\dfrac{1+9+9^2+...+9^{2010}}{1+9+9^2+...+9^{2009}}\)
B=\(\dfrac{1+5+5^2+...+5^{2010}}{1+5+5^2+...+5^{2009}}\)
Hãy so sánh A và B
Cho biểu thức: A =\(\frac{1+9+9^2+...+9^{2010}}{1+9+9^2+...+9^{2009}}\) B =\(\frac{1+5+5^2+...+5^{2010}}{1+5+5^2+...+2^{2009}}\)
Hãy so sánh A và B ???
Đặt M = \(1+9+9^2+......+9^{2010}\)
\(9M=9+9^2+9^3+......+9^{2011}\)
\(9M-M=8M=9^{2011}-1\)
Đặt K = \(1+9+9^2+......+9^{2009}\)
\(9K=9+9^2+9^3+.....+9^{2010}\)
\(9K-K=8K=9^{2010}-1\)
\(\Rightarrow A=\frac{9^{2011}-1}{9^{2010}-1}\)
Đặt H=\(1+5+5^2+....+5^{2010}\)
\(5H=5+5^2+......+5^{2011}\)
\(5H-H=4H=5^{2011}-1\)
ĐẶT G = \(1+5+5^2+.......+5^{2009}\)
\(5G-G=4G=5^{2010}-1\)
\(\Rightarrow B=\frac{5^{2011}-1}{5^{2010}-1}\)
Rồi bạn so sánh sẽ ra ngay
so sánh A=1+9+9^2+...+9^2010/1+9+9^2+...+9^2009 và B=1+5+5^2+...+5^2010/1+5+5^2+...+5^2009
cho biểu thức :
A=\(\frac{1+9^2+9^3+...+9^{2010}}{1+9+9^2+...+9^{2009}}\)
B=\(\frac{1+5^1+...+5^{2010}}{1+5+5^2+...+5^{2009}}\)
so sánh A vàB
A = \(1+\frac{9^{2010}}{1+9+9^2+....+9^{2009}}\)= \(1+1:\frac{1+9+9^2+....+9^{2009}}{9^{2010}}\)= \(1+1:\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+\frac{1}{9^{2008}}+...+\frac{1}{9}\right)\)
B = \(1+\frac{5^{2010}}{1+5+5^2+....+5^{2009}}\)= \(1+1:\frac{1+5+5^2+...+5^{2009}}{5^{2010}}\)= \(1+1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)
Do \(\frac{1}{9^{2010}}
Cho:
A=1+9+92+93+...+92010/1+9+...+92009
B=1+5+52+...+52010/1+5+52+...+52009
SO Sánh A và B
So sánh
a)A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\)và B=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)
b)M=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)và N=\(\frac{2009^{2009}-2}{2009^{2010}-2}\)
c)P=\(\frac{1+5+5^2+5^3+...+5^{10}}{1+5+5^2+5^3+...+5^9}\)và Q=\(\frac{1+3+3^2+3^3+...+3^{10}}{1+3+3^2+3^3+...+3^9}\)
a,Ta co:\(A=\frac{2005^{2005}+1}{2005^{2006}+1}<\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}\)
\(=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\) =B Vay A<B
b,lam tuong tu nhu y a
Bài 1:
a. \(\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
b. \(\left(-\dfrac{7}{45}\right)-\left(-\dfrac{1}{4}\right)-\left(-\dfrac{3}{5}\right)+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}-\left(-\dfrac{5}{9}\right)\)
\(a)\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
\(=\dfrac{5^{2007}\left(5^3+5^5+5^7\right)}{5^{2007}\left(5^4+5^2+1\right)}=\dfrac{5^3+5^5+5^7}{5^4+5^2+1}\)
\(=\dfrac{125+3125+78125}{625+25+1}=\dfrac{81375}{651}=125\)
\(b)-\dfrac{7}{45}+\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}+\dfrac{5}{9}\)
\(=\dfrac{-7.52+1.585+3.468+1.195+2.780+1.60-5.260}{2340}\)
\(=\dfrac{-364+585+1404+195+1560+60-1300}{2340}\)
\(=\dfrac{2140}{2340}=\dfrac{107}{117}\)
Tìm giá trị lớn nhất , giá trị nhỏ nhất của biểu thức :
a)\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\)
b)B=\(\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\)
c)C=\(-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ai lm đc câu nào thì giúp mk với , cảm ơn !!
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)
1,
a,tính:\(\dfrac{\dfrac{7}{2012}+\dfrac{7}{9}-\dfrac{1}{4}}{\dfrac{5}{9}-\dfrac{1}{2012}-\dfrac{1}{2}}\)
b,so sánh:A=\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2010};B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{17}\)
A=1-2+3+4-5-6+7+8-9-...+2007+2008-2009-2010
B=-1-2+3+4-5-6+7+8-9-10+11+12-...-2013-2014+2015+2016
C=(2+4+6+8+...+2014)-(3+5+7+9...+2011)
A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2005 + 2006 - 2007 - 2008 + 2009 + 2010 ( có 2010 số )
A = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + .... + ( 2005 + 2006 - 2007 - 2008 ) + ( 2009 + 2010 )
A = ( - 4 ) + ( - 4 ) + ... + ( - 4 ) + 4019 ( có 503 số )
A = ( - 4 ) . 502 + 4019
A = - 2008 + 4019
A = 2011.
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