b) Ta có: \(A=\dfrac{1012+1}{1013+1}\)

\(\Leftrightarrow A-1=\dfrac{1012+1-1013-1}{1013+1}\)

\(\Leftrightarrow A-1=\dfrac{-1}{1013+1}\)

Ta có: \(B=\dfrac{1011+1}{1012+1}\)

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

\(\Leftrightarrow B-1=\dfrac{-1}{1012+1}\)

Ta có: \(1013+1>1012+1\)

\(\Leftrightarrow\dfrac{1}{1013+1}< \dfrac{1}{1012+1}\)

\(\Leftrightarrow\dfrac{-1}{1013+1}>\dfrac{-1}{1012+1}\)

\(\Leftrightarrow A-1>B-1\)

hay A>B

Vậy: A>B

so sánh phải ko bn

Ta có : Q=\(\frac{1010+1011+1012}{1011+1012+1013}\)=\(\frac{1010}{1011+1012+1013}+\frac{1011}{1011+1012+1013}+\frac{1012}{1011+1012+1013}\)

Vì1010/1011>1010/1011+1012+1013

    1011/1012>1011/1011+1012+1013

    1012/1013>1012/1011+1012+1013

    =>P>Q

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

Sửa lại đề tý: \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\) mới có thể tính được nhé!

Ta có: \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(\Rightarrow A=1-\frac{1}{2020}=\frac{2020}{2020}-\frac{1}{2020}=\frac{2019}{2020}\)

Đến đây bạn tự làm tiếp nhé! Phân tích đến đây là dễ r =)

đề là như vậy bạn à ban đầu mk cũng nghĩ là sai đề nhg ko phải tại vì là đề thi HSG

Ta có :\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}=k\left(\frac{1}{1011}+\frac{1}{1012}+\frac{1}{1013}+....+\frac{1}{2020}\right)\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{2019}-\frac{1}{2020}=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2020}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2020}\right)=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2020}-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1010}=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

\(\Rightarrow\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2020}=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

=> k = 1

=> k là số tự nhiên (đpcm)

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

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

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

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!