Question

So sánh 

\(A=\frac{10^{1990}+1}{10^{1991}+1}\)

\(B=\frac{10^{1991}+1}{10^{1992}+1}\)

Answer 1

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 

Answer 2

A > B nha

Answer 3

10A=10^1991+10/10^1991+1       ;10B=10^1992+10/10^1992+1

10A=1+(10^1991+10-10^1991-1/10^1991+1)         ;10B=1+(10^1992+10-10^1992-1/10^1992+1)

10A=1+(9/10^1991+1)                                   ; 10B=1+(9/10^1992+1)

Có: 9/10^1991+1   >   9/10^1992+1

=>10A>10B

=>A>B