xet bt A ta co

A=2016.2017+1/2016.2017

=1+1/2016.2017

xet bt B ta co

B=2017.2018+1/2017.2018

=1+1/2017.2018

vì 1/2016.2017>1/2017.2018

nen 1+1/2016.2017>1+1/2017.2018

suy ra A>B

ai thay mik lam đúng thì k cho mik nha

b lớn hơn

Ta có:

\(A=\frac{2016.2017+1}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{1}{2016.2017}=1+\frac{1}{2016.2017}\)

\(B=\frac{2017.2018+1}{2017.2018}=\frac{2017.2018}{2017.2018}+\frac{1}{2017.2018}=1+\frac{1}{2017.2018}\)

So sánh:

\(A=\frac{1}{2016.2017}+1\)và \(B=\frac{1}{2017.2018}+1\)

Vì A và B đều cộng cho 1 nên tiêu giảm

=> \(A=\frac{1}{2016.2017}\)và \(B=\frac{1}{2017.2018}\)

Ta thấy:

Phần tử đã bằng nhau nên ta so sánh phần mẫu.

Ta thấy : 2016.2017 < 2017.2018

=>\(\frac{1}{2016.2017}>\frac{1}{2017.2018}\)

=> A > B

A=\(\dfrac{2016.2017+1}{2016.2017}=\dfrac{2016.2017}{2016.2017}+\dfrac{1}{2016.2017}=1+\dfrac{1}{2016.2017}\)

A=\(\dfrac{2017.2018+1}{2017.2018}=\dfrac{2017.2018}{2017.2018}+\dfrac{1}{2017.2018}=1+\dfrac{1}{2017.2018}\)

Mà 1=1; \(\dfrac{1}{2016.2017}\)>\(\dfrac{1}{2017.2018}\) nên A>B

\(\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

Ta thấy      \(2017.2018< 2018.2019\)

nên      \(\frac{1}{2017.1018}>\frac{1}{2018.2019}\)

\(\Rightarrow\)\(1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

Vậy      \(\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)

555555555555500000000000000.................

Ta có : \(\frac{2017.2018+1}{2017.2018}=1+\frac{1}{2017.2018}\)

             \(\frac{2018.2019+1}{2018.2019}=1+\frac{1}{2018.2019}\)

Mà : \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\) => \(\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)

a) \(\frac{53}{57}=\frac{530}{570}\)

Ta có : 1 - \(\frac{530}{570}\)= \(\frac{40}{570}\) ;    1 - \(\frac{531}{571}=\frac{40}{571}\)

Vì \(\frac{40}{570}>\frac{40}{571}\) nên \(\frac{53}{57}< \frac{531}{571}\)

Ta có:

\(C=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(D=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

Mà ta có:

\(\frac{1}{2017.2018}>\frac{1}{2018.2019}\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\Rightarrow C< D\)