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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 \(10^{1993}+10>10^{1992}+10\)
\(\Rightarrow\frac{11}{10^{1993}+10}< \frac{11}{10^{1992}+10}\)
\(\Rightarrow1-\frac{11}{10^{1993}+10}>1-\frac{11}{10^{1992}+10}\)
\(\Leftrightarrow\frac{1}{10}B>\frac{1}{10}A\)
\(\Rightarrow B>A\)
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 ~
\(\Rightarrow\frac{A}{10}=\frac{10^{1992}+1}{10^{1992}+10}=\frac{10^{1992}+10-9}{10^{1992}+10}=1-\frac{9}{10\left(10^{1991}+1\right)}\)
\(\Rightarrow\frac{B}{10}=\frac{10^{1993}+1}{10^{1993}+10}=\frac{10^{1993}+10-9}{10^{1993}+10}=1-\frac{9}{10\left(10^{1992}+1\right)}\)
Vì \(1-\frac{9}{10\left(10^{1991}+1\right)}< 1-\frac{9}{10\left(10^{1992}+1\right)}\Rightarrow A< B\)
Ta có : \(B=\frac{10^{1993}+1}{10^{1992}+1}\)< \(\frac{10^{1993}+10}{10^{1992}+10}\) = \(\frac{10.\left(10^{1992}+1\right)}{10.\left(10^{1993}+1\right)}\) = \(\frac{10^{1992}+1}{10^{1991}+1}\) = A
=> B < A
Vậy B < A
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
\(\frac{A}{10}=\frac{10^{1992}+1}{10^{1992}+10}=\frac{\left(10^{1992}+10\right)-9}{10^{1992}+10}=1-\frac{9}{10^{1992}+10}\)
\(\frac{B}{10}=\frac{10^{1993}+1}{10^{1993}+10}=\frac{\left(10^{1993}+10\right)-9}{10^{1993}+10}=1-\frac{9}{10^{1993}+10}\)
Vì \(10^{1992}+10< 10^{1993}+10\) nên \(1+\frac{9}{10^{1993}+10}>1+\frac{9}{10^{1993}+10}\)
Do đó \(A>B\)
