So sánh :
a)\(\frac{2^{2010}+1}{2^{2007}+1}\) và \(\frac{2^{2012}+1}{2^{2009}+1}\)
b)\(\frac{3^{123}+1}{3^{125}+1}\) và \(\frac{3^{122}+1}{3^{124}+1}\)
so sánh A và B:
A=\(\frac{3^{123}+1}{3^{125}+1}\)
B=\(\frac{3^{122}+1}{3^{124}+1}\)
Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\)N*)
Ta có:
\(A=\frac{3^{123}+1}{3^{125}+1}< \frac{3^{123}+1+2}{3^{125}+1+2}\)
\(A< \frac{3^{123}+3}{3^{125}+3}\)
\(A< \frac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\)
\(A< \frac{3^{122}+1}{3^{124}+1}=B\)
=> A < B
\(9A=\frac{3^{125}+9}{3^{125}+1}=1+\frac{8}{3^{125}+1}\)
\(9B=\frac{3^{124}+9}{3^{124}+1}=1+\frac{8}{3^{124}+1}\)
Mà 3^125+1>3^124+1 =>\(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)
Nên A<B
9A=\(\frac{3^{125}+9}{3^{125}+1}\)=\(1+\frac{8}{3^{125}+1}\)
9B=\(\frac{3^{124}+9}{3^{124}+1}\)=\(1+\frac{8}{3^{124}+1}\)
Vì \(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)\(\Rightarrow9B>9A\)\(\Rightarrow B>A\)
Vậy B>A
So Sánh
a)\(A=3^0+3^1+3^2+....+3^{2016}\) và \(B=3^{2017}\)
b)\(A=\frac{2^{2016}+1}{2^{2008}+1}\)và \(B=\frac{2^{2012}+1}{2^{2009}+1}\)
c)\(A=\frac{3^{123}+1}{3^{125}+1}\)và \(B=\frac{3^{122}}{3^{124}+1}\)
d)\(A=2^{30}+3^{20}+4^{30}\)và \(B=3.24^{10}\)
e)\(A=5^{84}\)và \(B=3^{126}\) và \(C=2^{108}\)
GIÚP MÌNH VỚI NHA! ( ^ _ ^ )
\(A=1+3+3^2+3^3+...+3^{2016}\)
\(A=1+3\left(1+3^2+...+3^{2015}\right)\)
\(A=1+3\left(A-3^{2016}\right)\)
\(A=1+3A-3^{2017}\)
\(2A=3^{2017}-1\Rightarrow A=\frac{3^{2017}-1}{2}\)
\(A< B\)
So Sánh
a) \(A=3^0+3^1+3^2+........+3^{2016}\) và \(B=3^{2017}\)
b)\(A=\frac{2^{2016}+1}{2^{2008}+1}\)và \(B=\frac{2^{2012}+1}{2^{2009}+1}\)
c)\(A=\frac{3^{123}+1}{3^{125}+1}\)và \(B=\frac{3^{122}}{3^{124}+1}\)
d)\(A=5^{84}\) và \(B=3^{126}\)và \(C=2^{168}\)
AI ĐÚNG MÌNH SẼ KICK CHO, LÀM ƠN GIÚP MÌNH VỚI NHÉ....( ~ _ ~ )
So sánh P và Q biết : P = 2010/2011 + 2011/2012 + 2012/2013 và Q = 2010+2011+2012/ 2011 +2012+2013
Chứng tỏ N < 1 với N = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}
so sánh: A= \(\frac{3^{123}+1}{3^{125}+1}\)và B= \(\frac{3^{122}}{3^{124}+1}\)
các bn lm nhanh hộ mik, mik đang cần gấp
\(B=\frac{3^{122}}{3^{124}+1}=\frac{3^{123}}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+1}=A\)
Do đó \(A>B\).
So sánh
A= \(\frac{2008^3+1}{2008^2-2007}\) và B= \(\frac{2009^3-1}{2009-2010}\)
Dễ thấy: \(2008^3+1>0\); \(2008^2-2007>0\)
Nên \(\frac{2008^3+1}{2008^2-2007}>0\Leftrightarrow A>0\)
và \(2009-2010< 0\); \(2009^3-1>0\)
\(\Rightarrow\frac{2009^3-1}{2009-2010}< 0\Leftrightarrow B< 0\)
Vậy A > B
Bài 1:So Sánh:200920và 2009200910
Bài 2:Tính tỉ số \(\frac{A}{B}\), biết:
\(A=\frac{1}{2}\)+\(\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\)
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
Bài 1:
Ta có: 200920=(20092)10=403608110 ; 2009200910=2009200910
Vì 403608110< 2009200910 => 200920< 2009200910
Bài 1:
Ta có:\(2009^{20}\)=\(2009^{10}\).\(2009^{10}\)
\(20092009^{10}\)=(\(\left(2009.10001\right)^{10}=2009^{10}.10001^{10}\)
Vì 2009<10001\(\Rightarrow2009^{20}< 20092009^{10}\)
So sánh
A=\(\frac{2006}{2007}-\frac{2007}{2008}+\frac{2008}{2009}-\frac{2009}{2010}\)
B=\(-\frac{1}{2006.2007}-\frac{1}{2008.2009}\)
So sánh
B=\(-\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
A=\(-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^4}\)
1.Tính tổng
\(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)
2.Tìm x
\(5^x+5^{x+2}=650\)
3.CMR
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
4. Cho \(A=\frac{1}{2010}+\frac{2}{2009}+\frac{3}{2008}+...+\frac{2009}{2}+\frac{2010}{1}\)
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2010}+\frac{1}{2011}\)
So sánh A và B