Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{-9\cdot10+\left(-19\right)}{10^{2011}}=\dfrac{-28}{10^{2011}}\)
\(B=\dfrac{-9\cdot10-19}{10^{2011}}=\dfrac{-109}{10^{2011}}\)
=>A>B
tách phana số -19/...... thành -9/....+(-10)/..... là so sanh đc bn ơi
\(A=-\frac{9}{10^{2010}}-\frac{19}{10^{2011}}=-\frac{9}{10^{2010}}-\frac{10}{10^{2010}}+\frac{10}{10^{2010}}-\frac{9}{10^{2011}}-\frac{10}{10^{2011}}.\)
\(=-\frac{19}{10^{2010}}-\frac{9}{10^{2011}}+\frac{1}{10^{2009}}-\frac{1}{10^{2010}}=B+\frac{1}{10^{2009}}-\frac{1}{10^{2010}}\)
\(\Rightarrow A-B=\frac{1}{10^{2009}}-\frac{1}{10^{2010}}>0\Rightarrow A>B.\)
\(-A=\frac{9}{10^{2010}}+\frac{19}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{10}{10^{2011}}+\frac{9}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{1}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{10}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{1}{10^{2009}}+\frac{9}{10^{2011}}\)
Tương tự với B, ta có:
\(-B=\frac{9}{10^{2011}}+\frac{19}{10^{2010}}\)
\(-B=\frac{9}{10^{2011}}+\frac{10}{10^{2010}}+\frac{9}{10^{2010}}\)
\(-B=\frac{9}{10^{2010}}+\frac{1}{10^{2009}}+\frac{9}{10^{2010}}\)
Ta thấy -B > -A \(\Rightarrow\)A > B.
\(A=-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2011}}\right)\)
\(B=-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2010}}\right)\)
Do \(-\frac{9}{10^{2011}}=-\frac{9}{10^{2011}}\)VÀ \(-\frac{19}{10^{2011}}< -\frac{19}{10^{2010}}\)
\(\Rightarrow-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2011}}\right)>-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2010}}\right)\)
\(\Leftrightarrow A>B\)
