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Ta có:
10A=\(\frac{10\left(10^{2017}+1\right)}{10^{2018}+1}=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}+\frac{9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)
10B=\(\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1}{10^{2019}+1}+\frac{9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)
do 1=1 và \(\frac{9}{10^{2018}+1}>\frac{9}{10^{2019}+1}\)
\(\Rightarrow\)A>B
Vậy A>B
chúc bạn học tốt!
Ta có: \(B=\frac{10^2\left(10^{2017}+1\right)}{10^2\left(10^{2016}+1\right)}=\frac{10^{2019}+1+99}{10^{2018}+1+99}\)
Do phân số \(A=\frac{10^{2019}+1}{10^{2018}+1}>1\).Áp dụng BĐT \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\left(m>0\right)\).
Ta có: \(A=\frac{10^{2019}+1}{10^{2018}+1}>\frac{10^{2019}+1+99}{10^{2018}+1+99}=B\)
Vậy \(A>B\)
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 !!!!!!!!
a) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)
=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)
=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)
\(A=\frac{10^{2017}}{10^{2018+1}}=\frac{10^{2017}}{10^{2019}}=\frac{1}{10^2}\)
Tương Tự với \(B=\frac{1}{10^2}\)
\(\Rightarrow A=B\)
Ta có : \(A=\frac{10^{2016}+1}{10^{2017}+1}\)
Suy ra \(10A=\frac{10^{2017}+10}{10^{2017}+1}\)
Suy ra \(10A=1+\frac{9}{10^{2017}+1}\)
Ta lại có : \(B=\frac{10^{2017}+1}{10^{2018}+1}\)
Suy ra : \(10B=\frac{10^{2018}+10}{10^{2018}+1}\)
Suy ra : \(10B=1+\frac{9}{10^{2018}+1}\)
Vì \(\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\)
Nên \(1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
Suy ra \(10A>10B\)
Suy ra \(A>B\)
Ta có :
A = \(\frac{10^{2017}+1}{10^{2018}+1}\)< 1 => A < \(\frac{10^{2017}+1+9}{10^{2018}+1+9}\)= \(\frac{10^{2017}+10}{10^{2018}+10}\)= \(\frac{10^{2016}+1}{10^{2017}+1}\)= B
Vậy A < B
A<B. lời giải thích khó viết lắm nên bạn tự tìm cách làm nhé
ta có :
\(A=\frac{10^{2019}+1}{10^{2018}+1}=\frac{10^{2018}.10+1}{10^{2018}+1}=\frac{10}{10^{2018}+1}\)
\(B=\frac{10^{2018}+1}{10^{2017}+1}=\frac{10^{2017}.10+1}{10^{2017}+1}=\frac{10}{10^{2017}+1}\)
Do \(10^{2017}+1< 10^{2018}+1\Rightarrow\frac{10}{10^{2017}+1}>\frac{10}{10^{2018}+1}\)
\(\Rightarrow A< B\)