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a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A
Vậy A > B
b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A
Vậy A < B.
NHỚ K CHO MK VỚI NHÉ !!!!!!!!
giải:
Ta có:
\(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-9}{10^{2010}}+\frac{-9-10}{10^{2011}}=\frac{-9}{10^{2010}}+\frac{-9}{10^{2011}}+\frac{-10}{10^{2011}}\)
\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-9-10}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}+\frac{-10}{10^{2010}}\)
Vì \(\frac{10}{10^{2011}}< \frac{10}{10^{2010}}\rightarrow\frac{-10}{10^{2011}}>\frac{-10}{10^{2010}}\Rightarrow\frac{-9}{10^{2010}}+\frac{-9}{10^{2011}}+\frac{-10}{10^{2011}}>\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}+\frac{-10}{10^{2010}}\)
Vậy \(A>B\)( Bạn nhớ đọc kĩ lời giải nhé)
\(A-B=\frac{10}{10^{2010}}-\frac{10}{10^{2011}}=\frac{1}{10^{2009}}-\frac{1}{10^{2010}}>0\)
\(\Rightarrow A>B\)
\(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-9}{10^{2010}}+\frac{-9-10}{10^{2011}}=\frac{-9}{10^{2010}}+\frac{-9}{10^{2011}}+\frac{-10}{10^{2011}}\)
\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-9-10}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}+\frac{-10}{10^{2010}}\)
Vì \(\frac{-10}{10^{2011}}>\frac{-10}{10^{2010}}\rightarrow A>B\)
