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Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)
=> \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
=> \(B< \frac{10^{1991}+10}{10^{1992}+10}\)
=> \(B< \frac{10.\left(10^{1990}+1\right)}{10.\left(10^{1991}+1\right)}\)
=> \(B< \frac{10^{1990}+1}{10^{1991}+1}=A\)
=> B < A
\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+10}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+10}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
Vì \(10^{1991}< 10^{1992}\Rightarrow1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}\)
\(\Rightarrow\frac{10^{1990}+1}{10^{1991}+1}>\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow A>B\)
Ta có : \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
Mà : \(\frac{10^{1991}+1+9}{10^{1992}+1+9}=\frac{10^{1991}+10}{10^{1992}+10}\)
\(=\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)
\(=\frac{10^{1990}+1}{10^{1991}+1}\)
\(\Rightarrow B< A\)
Ta có :
A = \(\frac{10^{1990}+1}{10^{1991}+1}\)
10A = \(\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
10A = \(\frac{10^{1991}+10}{10^{1991}+1}\)
10A = \(\frac{10^{1991}+1+9}{10^{1991}+1}\)
10A = \(1+\frac{9}{10^{1991}+1}\left(1\right)\)
Ta lại có :
B = \(\frac{10^{1991}+1}{10^{1992}+1}\)
10B = \(\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
10B = \(\frac{10^{1992}+10}{10^{1992}+1}\)
10B = \(\frac{10^{1992}+1+9}{10^{1992}+1}\)
10B = \(1+\frac{9}{10^{1992}+1}\left(2\right)\)
Từ \(\left(1\right)va\left(2\right)\)
Ta có :\(1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}\)
\(\Rightarrow\)10A > 10B
\(\Rightarrow\)A > B
Ta có:
\(A=\left(\frac{10^{1990}+1}{10^{1991}+1}\right).\frac{10}{10}=\frac{10^{1991}+10}{10^{1992}+10}\)
Mình làm bằng cách tính phần bù:
Ta có:
\(1-A=1-\frac{10^{1991}+10}{10^{1992}+10}=\frac{10^{1992}+10}{10^{1992}+10}-\frac{10^{1991}+10}{10^{1992}+10}=\frac{10^{1992}-10^{1991}}{10^{1992}+10}\)
\(1-B=1-\frac{10^{1991}+1}{10^{1992}+1}=\frac{10^{1992}+1}{10^{1992}+1}-\frac{10^{1991}+1}{10^{1992}+1}=\frac{10^{1992}-10^{1991}}{10^{1992}+1}\)
Vì \(\frac{10^{1992}-10^{1991}}{10^{1992}+10}\frac{10^{1991}+1}{10^{1992}+1}\)
\(\Rightarrow A>B\)
Vì\(\frac{10^{1991}+1}{10^{1992}+1}\)<1
Nên\(\frac{10^{1991}+1}{10^{1992}+1}\)<\(\frac{10^{1991}+1+9}{10^{1992}+1+9}\)
Ta có: \(\frac{10^{1991}+1+9}{10^{1992}+1+9}\)=\(\frac{10^{1991}+10}{10^{1992}+10}\)=\(\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\frac{10^{1990}+1}{10^{1991}+1}\)
=>\(\frac{10^{1991}+1}{10^{1992}+1}\)<\(\frac{10^{1990}+1}{10^{1991}+1}\)
Vậy: B<A
Đặt \(A=\frac{10^{1990}+1}{10^{1991}+1}\)
\(\Rightarrow10A=\frac{10\cdot(10^{1990}+1)}{10^{1991}+1}\)
\(=\frac{10^{1991}+10}{10^{1991}+1}=\frac{10^{1991}+1+9}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
Đặt \(B=\frac{10^{1991}+1}{10^{1992}+1}\)
\(\Rightarrow10B=\frac{10\cdot(10^{1991}+1)}{10^{1992}+1}=\frac{10^{1992}+10}{10^{1992}+1}=\frac{10^{1992}+1+9}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
Tự so sánh được rồi -_-
\(A=\frac{10^{1990}+1}{10^{1991}+1}vàB=\frac{10^{1991}+1}{10^{1992}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}
\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+1+9}{10^{1991}+1}\Rightarrow10A=1+\frac{9}{10^{1991}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+1+9}{10^{1992}+1}\Rightarrow10B=1+\frac{9}{10^{1992}+1}\)
=> 10A > 10B
=> A>B
Ta có : \(A=\frac{10^{1990}+1}{10^{1991}+1}=>10A=\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
\(=>10A=\frac{10^{1991}+10}{10^{1991}+1}=\frac{\left(10^{1991}+1\right)+9}{10^{1991}+1}\)
\(=>10A=1+\frac{9}{10^{1991}+1}\)
Ta lại có : \(B=\frac{10^{1991}+1}{10^{1992}+1}=>10B=\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
Tương tự như A => \(10B=1+\frac{9}{10^{1992}+1}\)
Vì \(\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}=>10A>10B\)
\(=>A>B\)
A < B
Chắc thế
:)
:)