So sánh A và B biết:
\(A=\frac{10^{2011}+1}{10^{2012}+1};B=\frac{10^{2012}+1}{10^{2013}+1}\)
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Ta có: \(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-9}{10^{2010}}-\frac{9}{10^{2011}}-\frac{10}{10^{2011}}\)
\(=\frac{-9}{10^{2010}}-\frac{9}{10^{1011}}-\frac{1}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}\)
Ta thấy : \(\frac{10}{10^{2010}}< \frac{19}{10^{2010}}\Rightarrow\frac{-10}{10^{2010}}>\frac{-19}{10^{2010}}\)
\(\Rightarrow\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}>\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
Hay \(A>B\)
Vậy ...
Ta có: \(A=\frac{10^{18}+1}{10^{19}+1}>\frac{10.\left(10^{17}+1\right)}{10.\left(10^{18}+1\right)}=\frac{10^{17}+1}{10^{18}+1}\)
Vậy A < B
Vì \(\frac{10^{18}+1}{10^{19}+1}< 1\Rightarrow B=\frac{10^{18}+1}{10^{19}+1}< \frac{10^{18}+1+9}{10^{19}+1+9}\)
\(\Rightarrow B< \frac{10^{18}+10}{10^{19}+10}\)
\(\Rightarrow B< \frac{10\left(10^{17}+1\right)}{10\left(10^{18}+1\right)}\)
\(\Rightarrow B< \frac{10^{17}+1}{10^{18}+1}\)
\(\Rightarrow B< A\)
Vậy A > B.
\(A=\left(1-\frac{1}{2011}\right)-\left(1-\frac{1}{2012}\right)+\left(1-\frac{1}{2013}\right)-\left(1-\frac{1}{2014}\right)\)
\(=1-\frac{1}{2011}-1+\frac{1}{2012}+1-\frac{1}{2013}-1+\frac{1}{2014}\)
\(=\left(1-1+1-1\right)-\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}+\frac{1}{2014}\right)\)
còn lại bó tay @@
\(A=\frac{2010}{2011}-\frac{2011}{2012}+\frac{2012}{2013}-\frac{2013}{2014}\)
và
\(B=\frac{1}{2010.2011}-\frac{1}{2012.2013}\)
Ta có: \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
Nên \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)\(\Rightarrow A>B\)
So sánh: \(\frac{2010}{2011}+\frac{2011}{2012}\) với \(\frac{2010+2011}{2011+2012}\)
Bạn ơi cho mình hỏi. Đây có phải bài trog toán tuổi thơ ko?
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
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 ...........