$\left(1989.1990+3978\right):\left(1992.1991-3984\right)$

$=\left[1989.\left(1990+2\right)\right]:\left[1992\left(1991-2\right)\right]=\left(1989.1992\right):\left(1992.1989\right)=1$

\(\left(1989.1990+3978\right):\left(1992.1991-3984\right)\)

\(=\left[1989.\left(1990+2\right)\right]:\left[1992\left(1991-2\right)\right]=\left(1989.1992\right):\left(1992.1989\right)=1\)

1989 . 1990 + 3978 1992 . 1991 − 3984  ⇒  1989 . 1990 + 1989 . 2 1992 . 1991 − 1992 . 2  ⇒  1989 . ( 1990 + 2 ) 1992 . ( 1991 + 2 )    ⇒  1989 . 1992 1992 . 1989  ⇒  1

dấu chấm là dấu nhân

\(\frac{1991x1993-1}{1990+1991x1992}=\frac{1991x\left(1992+1\right)-1}{1990+1991x1992}=\frac{1991x1992+1991-1}{1990+1991x1992}=\frac{1991x1992+1990}{1990+1991x1992}=1\)

\(\frac{1991.1993-1}{1990+1991.1992}=1\)

k nha

Kết quả là: \(\frac{1}{1990}\)

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

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 

Bài này còn có cách khác 

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

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