ta có:
1/10.A=10100+1/10(1099+1)
1/10.A=10100+1/10100+10
1/10.A=1-(9/10100+10)
1/10.B=10101+1/10(10100+1)
1/10.B=10101+1/10101+10
1/10.B=1-(9/10101+10)
vì(10101+10)>(10100+1)=> 9/10101+10 < 9/10100+10 => 1-(9/10101+10) > 1-(9/10100+10)
hay 1/10.A>1/10.B
=>A>B
ta có:
1/10.A=10100+1/10(1099+1)
1/10.A=10100+1/10100+10
1/10.A=1-(9/10100+10)
1/10.B=10101+1/10(10100+1)
1/10.B=10101+1/10101+10
1/10.B=1-(9/10101+10)
vì(10101+10)>(10100+1)=> 9/10101+10 < 9/10100+10 => 1-(9/10101+10) < 1-(9/10100+10)
hay 1/10.A<1/10.B
=>A<B
Đáp án dưới mới đúng nhé
vừa mình làm nhầm
\(A=\dfrac{10^{100}+1}{10^{99}+1}\)
\(\dfrac{1}{10}A=\dfrac{10^{100}+1}{10^{100}+10}=\dfrac{10^{100}+10-9}{10^{100}+10}\)
\(\dfrac{1}{10}A=\dfrac{10^{100}+10}{10^{100}+10}-\dfrac{9}{10^{100}+10}\)
\(\dfrac{1}{10}A=1-\dfrac{9}{10^{100}+10}\)
\(B=\dfrac{10^{101}+1}{10^{100}+1}\)
\(\dfrac{1}{10}B=\dfrac{10^{101}+1}{10^{101}+10}=\dfrac{10^{101}+10-9}{10^{101}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{101}+10}{10^{101}+10}-\dfrac{9}{10^{101}+10}\)
\(\dfrac{1}{10}B=1-\dfrac{9}{10^{101}+10}\)
Vì \(10^{100}+10< 10^{101}+10\) nên \(\dfrac{9}{10^{100}+10}>\dfrac{9}{10^{101}+10}\)
Vậy \(1-\dfrac{9}{10^{100}+10}< 1-\dfrac{9}{10^{101}+100}\) hay \(A< B\)