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\(A=\dfrac{-9\cdot10+\left(-19\right)}{10^{2011}}=\dfrac{-28}{10^{2011}}\)
\(B=\dfrac{-9\cdot10-19}{10^{2011}}=\dfrac{-109}{10^{2011}}\)
=>A>B
\(A=-\frac{9}{10^{2010}}-\frac{19}{10^{2011}}=-\frac{9}{10^{2010}}-\frac{10}{10^{2010}}+\frac{10}{10^{2010}}-\frac{9}{10^{2011}}-\frac{10}{10^{2011}}.\)
\(=-\frac{19}{10^{2010}}-\frac{9}{10^{2011}}+\frac{1}{10^{2009}}-\frac{1}{10^{2010}}=B+\frac{1}{10^{2009}}-\frac{1}{10^{2010}}\)
\(\Rightarrow A-B=\frac{1}{10^{2009}}-\frac{1}{10^{2010}}>0\Rightarrow A>B.\)
\(-A=\frac{9}{10^{2010}}+\frac{19}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{10}{10^{2011}}+\frac{9}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{1}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{10}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{1}{10^{2009}}+\frac{9}{10^{2011}}\)
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.
\(A-B=\frac{10}{10^{2010}}-\frac{10}{10^{2011}}=\frac{1}{10^{2009}}-\frac{1}{10^{2010}}>0\)
\(\Rightarrow A>B\)
\(A=-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2011}}\right)\)
\(B=-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2010}}\right)\)
Do \(-\frac{9}{10^{2011}}=-\frac{9}{10^{2011}}\)VÀ \(-\frac{19}{10^{2011}}< -\frac{19}{10^{2010}}\)
\(\Rightarrow-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2011}}\right)>-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2010}}\right)\)
\(\Leftrightarrow A>B\)
tách phana số -19/...... thành -9/....+(-10)/..... là so sanh đc bn ơi
A= -9/ 10 ^2010 + -19/ 10^2011=-9/ 10 ^2010 + -9/ 10^2011+-10/ 10^2011
B= -9/ 10^2011 + -19/ 10^2010= -9/ 10^2011 + -9/ 10 ^2010+-10/ 10^2010
vì -9/ 10^2011=-9/ 10^2011 và -9/ 10 ^2010=-9/ 10 ^2010
mà -10/ 10^2011>-10/ 10^2010 => A>B