Ta có :

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

\(\Rightarrow\frac{1}{10}A=\frac{10^{1992}+1}{10^{1992}+10}=\frac{10^{1992}+10-11}{10^{1992}+10}=1-\frac{11}{10^{1992}+10}\)

\(B=\frac{10^{1993}+1}{10^{1992}+1}\)

\(\Rightarrow\frac{1}{10}B=\frac{10^{1993}+1}{10^{1993}+10}=\frac{10^{1993}+10-11}{10^{1993}+10}=1-\frac{11}{10^{1993}+10}\)

Mà \(10^{1993}+10>10^{1992}+10\)

\(\Rightarrow\frac{11}{10^{1993}+10}< \frac{11}{10^{1992}+10}\)

\(\Rightarrow1-\frac{11}{10^{1993}+10}>1-\frac{11}{10^{1992}+10}\)

\(\Leftrightarrow\frac{1}{10}B>\frac{1}{10}A\)

\(\Rightarrow B>A\)

B > A k minh di co gi vao kb roi minh giai ki cho

A>B mình đảm bảo luôn

Ta có công thức : 

\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

\(\Rightarrow\)\(B>A\) hay \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~

Ta thấy \(10^{1993}+1>10^{1992}+1\)

\(\Rightarrow B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10.\left(10^{1992}+1\right)}{10.\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

\(\Rightarrow A< B\)

\(\frac{10^{1993}+1}{10^{1992}+1}>1\)

\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}\)

\(\frac{10^{1993+1}}{10^{1992}+1}>\frac{10^{1993}+10}{10^{1992}+10}\)

\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1992}+1}{10^{1991}+1}\)

thank you.cảm ơn bạn nhưng mình thấy bài giảng này không chắc lắm!

Có :

A = 10 - 9/10^1991+1

B = 10 - 9/10^1992+1

Vì 10^1991+1 < 10^1992+1 => 9/10^1991+1 > 9/10^1992+1

=> A < B

Tk mk nha

\(10A=\frac{10^{1993}+10}{10^{1993}+1}=1+\frac{9}{10^{1993}+1}\)

\(10B=\frac{10^{1994}+10}{10^{1994}+1}=1+\frac{9}{10^{1994}+1}\)

\(10^{1993}+1< 10^{1994}+1\Rightarrow\frac{9}{10^{1993}+1}>\frac{9}{10^{1994}+1}\)

\(\Rightarrow10A>10B\)

\(\Rightarrow A>B\)

Ta có B=\(\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}\)

        =  \(\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

   =>      B > A

bạn Trần Thùy Dung sai rồi A và B nhân với 1:10

\(\frac{A}{10}=\frac{10^{1992}+1}{10^{1992}+10}=\frac{\left(10^{1992}+10\right)-9}{10^{1992}+10}=1-\frac{9}{10^{1992}+10}\)

\(\frac{B}{10}=\frac{10^{1993}+1}{10^{1993}+10}=\frac{\left(10^{1993}+10\right)-9}{10^{1993}+10}=1-\frac{9}{10^{1993}+10}\)

Vì \(10^{1992}+10< 10^{1993}+10\) nên \(1+\frac{9}{10^{1993}+10}>1+\frac{9}{10^{1993}+10}\)

Do đó \(A>B\)

lấy máy tính mà tính!

tất nhiên A>B