các bạn giúp mình với nhanh lên nhé.Mình sẽ cho

A>B

BẠN Ạ

\(A=\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}\)

\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)

\(\frac{-9}{10^{2010}}>\frac{-19}{10^{2010}}\)

\(\Rightarrow\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}>\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)

\(\Rightarrow A>B\)

a,>

b,>

c,>

d,>

e,>

g,>

Ta có: \(A=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\)

\(=1+\left(\frac12+\frac13+\frac16\right)+\left(\frac14+\frac15\right)+\left(\frac17+\frac18+\frac19\right)+\frac{1}{10}\)

\(=1+1+\frac{9}{20}+\frac{1}{10}+\left(\frac17+\frac18+\frac19\right)\)

=2,55+\(\left(\frac17+\frac18+\frac19\right)\) >2,3

=>A>B

mọi người giúp mk với nhé

\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1=B\)

a/

\(\frac{2001}{2004}=\frac{2004-3}{2004}=1-\frac{3}{2004}=1-\frac{1}{668}.\)

\(\frac{39}{40}=\frac{40-1}{40}=1-\frac{1}{40}\)

Ta có \(40< 668\Rightarrow\frac{1}{40}>\frac{1}{668}\Rightarrow1-\frac{1}{40}< 1-\frac{1}{668}\Rightarrow\frac{39}{40}< \frac{2001}{2004}\)

b/

\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1=B\)

a)1024^9=(2^10)^9=2^90

2^100>2^90

b)125^80>20^118

c)9^12=(3^2)^12=3^24

27^10=(3^3)^10=3^30

3^24<3^30=9^12<27^10

tích nha