Ta có: \(\frac{2012}{2011}=1-\frac{1}{2011}\)

          \(\frac{2011}{2010}=1-\frac{1}{2010}\)

Vì \(\frac{1}{2011}>\frac{1}{2010}\)nên \(\frac{2012}{2011}>\frac{2011}{2010}\)

\(\Rightarrow\frac{2012}{2011}>\frac{2011}{2010}\)

\(\frac{2012}{2011}=1+\frac{1}{2011}\)
\(\frac{2011}{2010}=1+\frac{1}{2010}\)
vì  \(\frac{1}{2011}< \frac{1}{2010}\)
nên \(\frac{2012}{2011}< \frac{2011}{2010}\)

Ta có: 2012/2011 - 1 = 1/2011

          2011/2010 - 1 = 1/2010

Vì 1/2011 <1/2010 => 2012/2011 > 2011/2010.

\(\dfrac{2010}{2011}\) và \(\dfrac{2011}{2012}\)

Ta có:

\(1-\dfrac{2010}{2011}=\dfrac{1}{2011}\)

\(1-\dfrac{2011}{2012}=\dfrac{1}{2012}\)

Vì \(\dfrac{1}{2011}>\dfrac{1}{2012}\) nên \(\dfrac{2010}{2011}< \dfrac{2011}{2012}\)

20112010​ và 2011201220122011​

Ta có:

1−20102011=120111−20112010​=20111​

1−20112012=120121−20122011​=20121​

Vì 12011>1201220111​>20121​ nên 20102011<2011201220112010​<20122011​

ta có: 1 - 2010/2011= 2011/2011 - 2010/2011=1/2011

           1 - 2011/2012= 2012/2012 - 2011/2012= 1/2012

   Vì 1/2011>1/2012

   Nên 2010/2011>2011/2012

+ta có 10^2010=10...0(2010 số 0)

và 10^2011=10...0(2011 số 0)

suy ra  -9/10...0(2010 số 0)= -90/10...0(2011 số 0)[nhân tử,mẫu cho 10]

suy ra A=-90/10...0(2011 số 0)+-19/10...0(2011 số 0)= -109/10...0(2011 số 0)     [1]

+-19/10...0(2010 số 0)= -190/10...0(2011 số 0)[nhân tử,mẫu cho 10]

và 10^2011=10...0(2011 số 0)

suy ra -9/10...0(2011 số 0)+-190/10...0(2011 số 0)= -199/10...0(2011 số 0)    [2]

vì -109>-199 suy ra [1]>[2]

K CHO MIK VS BẠN ƠIIIIIIIIIIIIIIIIIII

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

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

\(-A=\frac{9}{10^{2010}}+\frac{1}{10^{2010}}+\frac{9}{10^{2011}}\)

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

\(-A=\frac{1}{10^{2009}}+\frac{9}{10^{2011}}\)

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

Làm tương tự nhé 

ta thấy -b > -a nên a>b

b,Ta có 

\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)

\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)

\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)

\(\Rightarrow P>Q\)

\(A=\frac{-10}{20}+\frac{-10}{30}+\frac{-10}{42}+\frac{-10}{56}+\frac{-10}{72}+\frac{-10}{90}+\frac{-10}{110}\)

\(=-10\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)

\(=-10\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

\(=-10\left(\frac{1}{4}-\frac{1}{11}\right)\)

\(=\frac{-35}{22}\)

giải:

Ta có:

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

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

Vì \(\frac{10}{10^{2011}}< \frac{10}{10^{2010}}\rightarrow\frac{-10}{10^{2011}}>\frac{-10}{10^{2010}}\Rightarrow\frac{-9}{10^{2010}}+\frac{-9}{10^{2011}}+\frac{-10}{10^{2011}}>\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}+\frac{-10}{10^{2010}}\)

Vậy \(A>B\)( Bạn nhớ đọc kĩ lời giải nhé)

 −910²⁰¹⁰ +−1910²⁰¹¹  = −910²⁰¹¹ +−1910²⁰¹⁰