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

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

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!

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!

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

\(C=\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\right)-\)\(\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}\right)\)

\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+..+\frac{1}{2017}\)

\(\Rightarrow C=\left(\frac{1}{101}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\right)-\left(\frac{1}{1010}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)

\(\Rightarrow C=\frac{1}{2018}\)

Bạn có thể viết lại đề theo phân số như thế này được không \(\frac{7}{12}\)bạn viết thế mk ko hiểu

Bn viết lại đề nhanh mk làm cho

Chúc bn học tốt

i am Chịu!!!!!

thích vênh thì tự đi mà làm, nhóc con đấy thì sao ạ