Giải:

Ta có: A=10¹¹-1/10¹²-1

       10A=10.(10¹¹-1)/10¹²-1

       10A=10¹²-10/10¹²-1

       10A=10¹²-1-9/10¹²-1

       10A=10¹²-1/10¹²-1 - 9/10¹²-1

       10A=1-9/10¹²-1

Tương tự: B=10¹⁰+1/10¹¹+1

              10B=1+9/10¹¹+1

Vì -9/10¹²-1 < 9/10¹¹+1 nên 10A < 10B

Vậy A<B

Chúc bạn học tốt!

Giải:

A=10^11-1/10^12-1

10A=10.(10^11-1)/10^12-1

10A=10^12-10/10^12-1

10A=10^12-1-9/10^12-1

10A=10^12-1/10^12-1 + -9/10^12-1

10A=1+ -9/10^12-1

B=10^10+1/10^11+1

10B=10.(10^10+1)/10^11+1

10B=10^11+10/10^11+1

10B=10^11+1+9/10^11+1

10B=10^11+1/10^11+1 + 9/10^11+1

10B=1 + 9/10^11+1

Vì -9/10^12-1 < 9/10^11+1 nên 10A < 10B

=>A < B

Chúc bạn học tốt!

\(A=\dfrac{1011-1}{1012-1}=\dfrac{1010}{1011}\)

\(B=\dfrac{1010+1}{1011+1}=\dfrac{1011}{1012}\)

Ta có :

\(1-A=1-\dfrac{1010}{1011}=\dfrac{1}{1011}\)

\(1-B=1-\dfrac{1011}{1012}=\dfrac{1}{1012}\)

NHận thấy \(\dfrac{1}{1011}>\dfrac{1}{1012}\Rightarrow A< B\)

Ta có:

\(A=\dfrac{1011-1}{1012-1}=\dfrac{1010}{1011}\)

\(B=\dfrac{1010+1}{1011+1}=\dfrac{1011}{1012}\)

Ta lại có:

\(1-\dfrac{1010}{1011}=\dfrac{1}{1011}\)

\(1-\dfrac{1011}{1012}=\dfrac{1}{1012}\)

Vì \(\dfrac{1}{1011}>\dfrac{1}{1012}\Rightarrow\dfrac{1010}{1011}< \dfrac{1011}{1012}\Rightarrow A< B\)

Ta có: \(A-B\)

\(=1011\left(1+\frac13+\frac15+\cdots+\frac{1}{2019}\right)-1010\left(\frac12+\frac14+\cdots+\frac{1}{2020}\right)\)

\(=1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)+\left(1+\frac13+\cdots+\frac{1}{2019}\right)\)

Vì \(1-\frac12>0;\frac13-\frac14>0;\ldots;\frac{1}{2019}-\frac{1}{2020}>0\)

nên \(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}>0\)

=>\(1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)>0\)

=>A-B>0

=>A>B

dễ lắm hải ơi

em vừa hỏi cô Huyền xong , đơn giản cực

Ta có: \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2021\cdot2022}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2022}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\)

Ta có: \(B=1011+\frac{1010}{1012}+\frac{1009}{1013}+\cdots+\frac{2}{2020}+\frac{1}{2021}\)

\(=\left(\frac{1010}{1012}+1\right)+\left(\frac{1009}{1013}+1\right)+\cdots+\left(\frac{2}{2020}+1\right)+\left(\frac{1}{2021}+1\right)+1\)

\(=\frac{2022}{1012}+\frac{2022}{1013}+\cdots+\frac{2022}{2022}=2022\left(\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\right)\)

=2022A

=>\(\frac{B}{A}=2022\) là số nguyên