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15 tháng 1 2017

A < B nha!

2 tháng 3 2017

Vì \(\frac{10^{2011}+1}{10^{2012}+1}< 1\)

=> \(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}=\frac{10^{2011}+10}{10^{2012}+10}=\frac{10\left(10^{2010}+1\right)}{10\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)

Vậy A > B

2 tháng 3 2017

A>B hay sao y

13 tháng 1 2018

\(B< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)

Vậy A > B

13 tháng 1 2018

Áp dụng bất đẳng thức :

\(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

Ta có :

\(B=\frac{10^{2012}+1}{10^{2013}+1}< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)

\(\Leftrightarrow B< A\)

5 tháng 7 2017

a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)

Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)

\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)

Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)

Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)

b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)

Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)

Vậy A > B 

Có gì  sai cho sorry

a,

\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)

b,

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

1 tháng 7 2016

Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
\(=>B=\frac{10^{2012}+1}{10^{2013}+1}< \frac{10^{2012}+1+9}{10^{2013}+1+9}\)

                                          \(< \frac{10^{2012}+10}{10^{2013}+10}\)

                                          \(< \frac{10.\left(10^{2011}+1\right)}{10.\left(10^{2012}+1\right)}\)

                                          \(< \frac{10^{2011}+1}{10^{2012}+1}=A\)

=> B < A

Ủng hộ mk nha ^_-

9 tháng 3 2015

Cho C=\(10^{2010}+\frac{1}{10^{2010}}\)

Xét \(A_1=10^{2010}+\frac{1}{10^{2011}}\)và \(B^{ }_1=10^{2011}+\frac{1}{10^{2012}}\)

Ta có \(A_1-C=10^{2010}+\frac{1}{10^{2010}}-10^{2010}-\frac{1}{10^{2010}}\)

         \(A_1-C=10.\left(\frac{1}{10^{2011}}-\frac{1}{10^{2010}}\right)\)

Giair tượng tự ta được \(B_1-C=10^{2010}.\left(9+\frac{1}{10^{2012}}-\frac{1}{10^{2010}}\right)\)

Ta thấy \(\frac{1}{10^{2012}}-\frac{1}{10^{2010}}

3 tháng 1 2020

Có \(\hept{\begin{cases}A=\frac{-9}{10^{2012}}+\frac{-19}{10^{2011}}\\B=\frac{-19}{10^{2012}}+\frac{-9}{10^{2011}}\end{cases}}\)

\(\Rightarrow\)A-B=\(\frac{10}{10^{2011}}-\frac{10}{10^{2012}}=\frac{1}{10^{2010}}-\frac{1}{10^{2011}}>0\)

\(\Rightarrow A>B\)

30 tháng 6 2018

a) Ta có :

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

\(\Rightarrow10A=\frac{10^{2011}+10}{10^{2011}+1}=\frac{\left(10^{2011}+1\right)+9}{10^{2011}+1}=1+\frac{9}{10^{2011}+1}\)

\(B=\frac{10^{2011}+1}{10^{2012}+1}\)

\(\Rightarrow10B=\frac{10^{2012}+10}{10^{2012}+1}=\frac{\left(10^{2012}+1\right)+9}{10^{2012}+1}=1+\frac{9}{10^{2012}+1}\)

Vì \(\frac{9}{10^{2011}+1}>\frac{9}{10^{2012}+1}\)nên \(10A>10B\)

\(\Rightarrow A>B\)

Vậy : \(A>B\)

b) Ta có :

\(\left(\frac{-1}{2}\right)^{11}=\frac{-1^{11}}{2^{11}}=\frac{-1}{2^{11}}\)

\(\left(\frac{-1}{2}\right)^{13}=\frac{-1^{13}}{2^{13}}=\frac{-1}{2^{13}}\)

Vì \(\frac{-1}{2^{11}}>\frac{-1}{2^{13}}\)nên \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)

Vậy : \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)

30 tháng 6 2018

\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}\)

\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+10}{10^{2012}+10}\)

\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10\cdot\left(10^{2010}+1\right)}{10\cdot\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)

Vậy : B < A