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Giải:
Ta gọi \(\dfrac{10^{1990}+1}{10^{1991}+1}\) =A và \(\dfrac{10^{1991}}{10^{1992}}\) =B
Ta có:
A=\(\dfrac{10^{1990}+1}{10^{1991}+1}\)
10A=\(\dfrac{10^{1991}+10}{10^{1991}+1}\)
10A=\(\dfrac{10^{1991}+1+9}{10^{1991}+1}\)
10A=\(1+\dfrac{9}{10^{1991}+1}\)
Tương tự:
B=\(\dfrac{10^{1991}}{10^{1992}}\)
10B=\(\dfrac{10^{1992}}{10^{1992}}=1\)
Vì \(\dfrac{9}{10^{1991}+1}< 1\) nên 10A<10B
⇒ \(\dfrac{10^{1990}+1}{10^{1991}+1}\) < \(\dfrac{10^{1991}}{10^{1992}}\)
Giải:
a) \(A=\dfrac{10^{1990}+1}{10^{1991}+1}\) và \(B=\dfrac{10^{1991}+1}{10^{1992}+1}\)
Ta có:
\(A=\dfrac{10^{1990}+1}{10^{1991}+1}\)
\(10A=\dfrac{10^{1991}+10}{10^{1991}+1}\)
\(10A=\dfrac{10^{1991}+1+9}{10^{1991}+1}\)
\(10A=1+\dfrac{9}{10^{1991}+1}\)
Tương tự :
\(B=\dfrac{10^{1991}+1}{10^{1992}+1}\)
\(10B=\dfrac{10^{1992}+10}{10^{1992}+1}\)
\(10B=\dfrac{10^{1992}+1+9}{10^{1992}+1}\)
\(10B=1+\dfrac{9}{10^{1992}+1}\)
Vì \(\dfrac{9}{10^{1991}+1}>\dfrac{9}{10^{1992}+1}\) nên \(10A>10B\)
\(\Rightarrow A>B\left(đpcm\right)\)
Chúc bạn học tốt!
Ta có :
\(10A=\dfrac{10\left(10^{1990}+1\right)}{10^{1991}+1}=\dfrac{10^{1991}+10}{10^{1991}+1}=\dfrac{10^{1991}+1+9}{10^{1991}+1}=1+\dfrac{9}{10^{1991}+1}\left(1\right)\)
\(10B=\dfrac{10\left(10^{1991}+1\right)}{10^{1992}+1}=\dfrac{10^{1992}+10}{10^{1992}+1}=\dfrac{10^{1992}+1+9}{10^{1992}+1}=1+\dfrac{9}{10^{1992}+1}\left(2\right)\)
Lại có : \(1+\dfrac{9}{10^{1991}+1}>1+\dfrac{9}{10^{1992}+1}\)
\(\Leftrightarrow10A>10B\Leftrightarrow A>B\)
Vậy...
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
Ta có :
\(10A=\dfrac{10^{1991}+10}{10^{1991}+1}=\dfrac{10^{1991}+1+9}{10^{1991}+1}=1+\dfrac{9}{10^{1991}+1}\)\(\left(1\right)\)
\(10B=\dfrac{10^{1992}+10}{10^{1992}+1}=\dfrac{10^{1992}+1+9}{10^{1992}+1}=1+\dfrac{9}{10^{1992}+1}\)\(\left(2\right)\)
Vì \(1+\dfrac{9}{10^{1991}+1}>1+\dfrac{9}{10^{1992}+1}\)\(\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
~ Chúc bn học tốt ~
Ta có:
A=101990+1101991+1=101990.10101991.10=101990101991=1/10A=101990+1101991+1=101990.10101991.10=101990101991=1/10 (%)
B=101991+1101992+1=101991.10101992.10=101991101992=1/10B=101991+1101992+1=101991.10101992.10=101991101992=1/10 (%) (%)