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 ???
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}}
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=\(\dfrac{1+9+9^2+...+9^{2010}}{1+9+9^2+...+9^{2009}}\)
B=\(\dfrac{1+5+5^2+...+5^{2010}}{1+5+5^2+...+5^{2009}}\)
Hãy so sánh A và B
Ta có :
+) \(A=\dfrac{1+9+9^2+...+9^{2009}}{1+9+9^2+...+9^{2009}}+\dfrac{9^{2010}}{1+9+9^2+...+9^{2009}}\)
\(A=1+1:\dfrac{1+9+9^2+...+9^{2009}}{9^{2010}}\)
\(A=1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)\)
+) \(B=\dfrac{1+5+5^2+...+5^{2009}}{1+5+5^2+...+5^{2009}}+\dfrac{5^{2010}}{1+5+5^2+...+5^{2009}}\)
\(B=1+1:\dfrac{1+5+5^2+...+5^{2009}}{5^{2010}}\)
\(B=1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
Vì \(\dfrac{1}{9^{2010}}< \dfrac{1}{5^{2010}}\)
\(\dfrac{1}{9^{2009}}< \dfrac{1}{5^{2009}}\) (ngoặc cả mấy cài so sánh này vào rôi mời suy ra nhé)
.............................
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\)=> \(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}< \dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\)
=> \(1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
=> \(1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
Hay A > B
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
So Sánh A vs B :
A=\(\frac{100^{2009}+1}{100^{2008}+1}\) B=\(\frac{100^{2010}+1}{100^{2009}+1}\)
A=\(\frac{5^0+5^1+5^2+5^3+.....+5^9}{5^0+5^1+5^2+.....+5^8}\) B=\(\frac{3^0+3^1+3^2+.....+3^9}{3^0+3^1+3^2+.....+3^8}\)
Giúp vsssssssssssssssssssssssssssssssssssssssss nhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa .........................
1.Cho 99 số nguyên trong đó tổng của 14 số bất kì là một số dương.Chứng tỏ rằng tổng của 99 số đó là số dương
2.a,So sánh A và B biết : A=\(\frac{2}{9^4}+\frac{7}{9^5}\) và B=\(\frac{7}{9^4}+\frac{2}{9^5}\)
b,Cho A=(-5)2+(-5)3+(-5)4+...+(-5)2014.A có chia hết cho 21 không? Vì sao?
3,Cho x=22010-22009-22008-...-2-1.Tính 2012x
a,A=1-2+3+4-5-6+7+8-9-...+2007+2008-2009-2010
b, \(\frac{1}{5^2}-\frac{1}{5^3}+\frac{1}{5^4}-\frac{1}{5^5}+..-\frac{1}{5^{101}}\).CM<\(\frac{1}{30}\)
Tính giá trị biểu thức
A=1-2+3+4-5-6+7+8-9-.....+2007+2008-2009-2010
B=\(1-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-...-\frac{1}{9900}\)
b: \(B=1-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}-\dfrac{49}{100}=\dfrac{1}{100}\)