MIK CHỊU CHƯA HỌC

=1/1990

mình chả biết đúng hay sai nữa

1991*1993-1/1990+1991*1992

=1991*(1992+1)-1/1990+1991*1992

=1991*1992+1991*1-1/1990+1991*1992

=1990/1990

=1

   1991*(1993-1)-1/1990+1991*1992

= 1991*1992-1/1990+1991*1992

= 1990/1990

= 1

\(\frac{1991.1993-1}{1992+1990.1993}=\frac{1990.1993+1993-1}{1992+1990.1993}=\frac{1992+1990.1993}{1992+1990.1993}=1\)

\(\frac{1993.1991-1}{1992+1990.1993}=\frac{1993.\left(1990+1\right)-1}{1992+1990.1993}=\frac{1993.1990+1993-1}{1992+1990.1993}=\frac{1993.1990+1992}{1992+1990.1993}=1\)

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 -_-

sao ra được 1+ gì gì đó vậy bạn

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

\(A=\frac{10^{1990}+1}{10^{1991}+1}vàB=\frac{10^{1991}+1}{10^{1992}+1}\)

\(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

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

Bài này mình biết làm nè , nhưng ... dài dòng lắm 

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 

A > B nha

\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+10}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)

\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+10}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)

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 : \(\frac{10^{1991}+1+9}{10^{1992}+1+9}=\frac{10^{1991}+10}{10^{1992}+10}\)

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

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

\(\Rightarrow B< A\)