Ta có : \(A=\frac{10^{1990}+1}{10^{1991}+1}=>10A=\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
\(=>10A=\frac{10^{1991}+10}{10^{1991}+1}=\frac{\left(10^{1991}+1\right)+9}{10^{1991}+1}\)
\(=>10A=1+\frac{9}{10^{1991}+1}\)
Ta lại có : \(B=\frac{10^{1991}+1}{10^{1992}+1}=>10B=\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
Tương tự như A => \(10B=1+\frac{9}{10^{1992}+1}\)
Vì \(\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}=>10A>10B\)
\(=>A>B\)
đăt 10A=\(\frac{10^{1991}+1}{10^{1991}+1}\)=1+\(\frac{9}{10^{1991}}\)
Câu B tương tự
ta có:\(\frac{9}{10^{1991}+1}\)>\(\frac{9}{10^{1992}}\)
nên 10A>10B
=>A>b
Ta có :
B \(=\frac{10^{1991}+1}{10^{1992}+1}\)\(< \frac{10^{1991}+1+9}{10^{1992}+1+9}=\frac{10^{1991}+10}{10^{1992}+10}=\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1990}+1}{10^{1991}+1}=A\)
vậy \(A>B\)
