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19 tháng 6 2015

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

=>10A=\(\frac{10\left(10^{2011}+10\right)}{10^{2012}+10}=\frac{10^{2012}+100}{10^{2012}+10}=\frac{10^{2012}+10}{10^{2012}+10}+\frac{90}{10^{2012}+10}=1+\frac{90}{10^{2012}+10}\)

đặt B=\(\frac{10^{2012}-10}{10^{2013}-10}\)

=>10B=\(\frac{10\left(10^{2012}-10\right)}{10^{2013}-10}=\frac{10^{2013}-100}{10^{2013}-10}=\frac{10^{2013}-10}{10^{2013}-10}+\frac{-90}{10^{2013}-10}=1+\frac{-90}{10^{2013}-10}\)

vì \(\frac{-90}{10^{2013}-10}\) luôn âm nên 

\(1+\frac{90}{10^{2012}+10}>1+\frac{-90}{10^{2013}-10}\)

vậy \(A>Bhay\frac{10^{2011}+10}{10^{2012}+10}>\frac{10^{2012}-10}{10^{2013}-10}\)

dễ mà cũng hỏi à mình học lớp 4 đó

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,

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

21 tháng 3 2020

Có : \(A=\frac{10^{2012}-10}{10^{2013}-10}\)

\(\Leftrightarrow10A=\frac{10^{2013}-100}{10^{2013}-10}\)

\(\Leftrightarrow10A=\frac{10^{2013}-10-90}{10^{2013}-10}\)

\(\Leftrightarrow10A=1-\frac{90}{10^{2013}-10}\)

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

\(\Leftrightarrow10B=\frac{10^{2012}+100}{10^{2012}+10}\)

\(\Leftrightarrow10B=\frac{10^{2012}+10+90}{10^{2012}+10}\)

\(\Leftrightarrow B=1+\frac{90}{10^{2012}+10}\)

Ta thấy : \(1-\frac{90}{10^{2013}-10}< 1\)

              \(1+\frac{90}{10^{2012}+10}>1\)

\(\Leftrightarrow1-\frac{90}{10^{2013}-10}< 1+\frac{90}{10^{2012}+10}\)

\(\Leftrightarrow A< B\)

15 tháng 1 2017

A < B nha!

12 tháng 1 2019

b,Ta có 

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

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

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

\(\Rightarrow P>Q\)

12 tháng 1 2019

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

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

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

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

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

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 ^_-

28 tháng 2 2016

So sánh 2 phân số sau  $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10 

kick dzô chữ xanh là được!! OK

28 tháng 2 2016

Ta có : 

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

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

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

         = \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)

         = 1 - \(\frac{9}{10^{2012}+1}\)

10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)

          = \(\frac{10^{2013}+10}{10^{2013}+1}\)

          = \(\frac{10^{2013}+1+9}{10^{2013}+1}\)

          = 1 - \(\frac{9}{10^{2013}+1}\)

Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\)  nên 10.A > 10.B

=> A >B 

Vậy ...........

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

\(\Rightarrow A>B\)