Không quy đồng mẫu hãy so sánh
A=-9/102010+-19/102011; B=-9/102011+-19/102010
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\(\dfrac{1}{10}A=\dfrac{10^{2012}+1}{10^{2012}+10}=1-\dfrac{9}{10^{2012}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{2011}+1}{10^{2011}+10}=1-\dfrac{9}{10^{2011}+10}\)
10^2012+10>10^2011+10
=>9/10^2012+10<9/10^2011+10
=>-9/10^2012+10>-9/10^2011+10
=>A>B
\(A=\dfrac{-9\cdot10+\left(-19\right)}{10^{2011}}=\dfrac{-28}{10^{2011}}\)
\(B=\dfrac{-9\cdot10-19}{10^{2011}}=\dfrac{-109}{10^{2011}}\)
=>A>B
Sửa đề: Chứng mình chia hết 24
Tách: 24=8.3
⇒3 (1)
8 (Vì: 0088) (2)
Từ (1) và (2) ⇒A24 Vì: (3,8)
⇒đpcm
\(A=-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2011}}\right)\)
\(B=-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2010}}\right)\)
Do \(-\frac{9}{10^{2011}}=-\frac{9}{10^{2011}}\)VÀ \(-\frac{19}{10^{2011}}< -\frac{19}{10^{2010}}\)
\(\Rightarrow-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2011}}\right)>-\frac{9}{10^{2011}}+\left(-\frac{19}{10^{2010}}\right)\)
\(\Leftrightarrow A>B\)
\(A=-\frac{9}{10^{2010}}-\frac{19}{10^{2011}}=-\frac{9}{10^{2010}}-\frac{10}{10^{2010}}+\frac{10}{10^{2010}}-\frac{9}{10^{2011}}-\frac{10}{10^{2011}}.\)
\(=-\frac{19}{10^{2010}}-\frac{9}{10^{2011}}+\frac{1}{10^{2009}}-\frac{1}{10^{2010}}=B+\frac{1}{10^{2009}}-\frac{1}{10^{2010}}\)
\(\Rightarrow A-B=\frac{1}{10^{2009}}-\frac{1}{10^{2010}}>0\Rightarrow A>B.\)
\(-A=\frac{9}{10^{2010}}+\frac{19}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{10}{10^{2011}}+\frac{9}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{1}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{10}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{1}{10^{2009}}+\frac{9}{10^{2011}}\)
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.