so sánh: A=(-9)/102019+(-19)/102020 và B=-9/102020+(-19)/102019
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(10A=\dfrac{10^{2021}+10}{10^{2021}+1}=\dfrac{\left(10^{2021}+1\right)+9}{10^{2021}+1}=\dfrac{10^{2021}+1}{10^{2021}+1}+\dfrac{9}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=\dfrac{\left(10^{2022}+1\right)+9}{10^{2022}+1}=\dfrac{10^{2022}+1}{10^{2022}+1}+\dfrac{9}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
Vì \(10^{2022}>10^{2021}=>10^{2021}+1< 10^{2022}+1\)
\(=>\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\)
\(=>10A>10B\)
\(=>A>B\)
So sánh:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}\) và \(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}\)
Ta có:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}=-\frac{1}{10^{2011}}\left(\frac{9}{10}+19\right)=-\frac{1}{10^{2011}}.\frac{199}{10}\)
\(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}=-\frac{1}{10^{2011}}\left(9+\frac{19}{10}\right)=-\frac{1}{10^{2011}}.\frac{109}{10}\)
Vì \(\frac{199}{10}>\frac{109}{10}\Rightarrow\frac{1}{10^{2011}}.\frac{199}{10}>\frac{1}{10^{2011}}.\frac{109}{10}\Rightarrow-\frac{1}{10^{2011}}.\frac{199}{10}< -\frac{1}{10^{2011}}.\frac{109}{10}\)
Vậy nên A<B
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 !!!!!!!!
ta có -9\10^2011=-9\10^2011
mà -19\10^2011>-19\10^2011
nên 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
A=\(\frac{-199}{10^{2011}}\)
B=\(\frac{-109}{10^{2011}}\)
Dễ dàng so sánh được A<B