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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: \(\hept{\begin{cases}A=\frac{10^{2016}+1}{10^{2017}+1}\\B=\frac{10^{2017}+1}{10^{2018}+1}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}10A=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\\10B=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\end{cases}}\)
Vì \(\frac{9}{10^{1017}+1}>\frac{9}{10^{2018}+1}\)
nên \(10A>10B\Rightarrow A>B\)
\(A=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\frac{10\cdot(10^{2016}+1)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)
\(A=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{2017}+1\)
\(\Rightarrow\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)
\(\Rightarrow\)\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)
....
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: 10A = \(\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
10B = \(\frac{10^{2018}+1+9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)
\(vì\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\) => 10A > 10B => A > B
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 \(A>B\)
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!
\(P=\frac{10^{2017}-1}{10^{2018}-1}\) \(>\) \(Q=\frac{10^{2016}+1}{10^{2017}+1}\)