Đặt C = 1 + 2017 + 2017² + ... + 2017²⁰¹⁶ ; D = 1 + 2016 + 2016² + ... + 2016²⁰¹⁶

Ta có : 2017C = 2017 + 2017² + 2017³ + ... + 2017²⁰¹⁷

=> 2016C = 2017C - C = 2017²⁰¹⁷ - 1\(\Rightarrow C=\frac{2017^{2017}-1}{2016}\)

2016D = 2016 + 2016² + 2016³ + ... + 2016²⁰¹⁷

=> 2015D = 2016D - D = 2016²⁰¹⁷ - 1\(\Rightarrow D=\frac{2016^{2017}-1}{2015}\)

\(\Rightarrow A=\frac{2017^{2017}}{\frac{2017^{2017}-1}{2016}}=\frac{2017^{2017}.2016}{2017^{2017}-1}\);\(B=\frac{2016^{2017}}{\frac{2016^{2017}-1}{2015}}=\frac{2016^{2017}.2015}{2016^{2017}-1}\)

Ta có : 2017²⁰¹⁷.2016.(2016²⁰¹⁷ - 1) - 2016²⁰¹⁷.2015.(2017²⁰¹⁷ - 1)

= 2017²⁰¹⁷.2016²⁰¹⁷.2016 - 2017²⁰¹⁷.2016 - 2017²⁰¹⁷.2016²⁰¹⁷.2015 + 2016²⁰¹⁷.2015

= 2017²⁰¹⁷.2016²⁰¹⁷ - 2017²⁰¹⁷.2016 + 2016²⁰¹⁷.2015

= 2017²⁰¹⁷.(2016²⁰¹⁷ - 2016) + 2016²⁰¹⁷.2015 > 0

=> A > B

Ta có 

\(A=1:\frac{1+2017+2017^2+...+2017^{2016}}{2017^{2017}}\)

\(B=1:\frac{1+2016+2016^2+...2016^{2016}}{2016^{2017}}\)

\(A=1:\left(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}\right)\)

\(B=1:\left(\frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\right)\)

Có 2017²⁰¹⁷>2016²⁰¹⁷ ;  2017²⁰¹⁶>2016²⁰¹⁶ ;  2017²⁰¹⁵>2016²⁰¹⁵;..... ; 2017>2016

=> \(\frac{1}{2017^{2017}}< \frac{1}{2016^{2017}};\frac{1}{2017^{2016}}< \frac{1}{2016^{2016}};\frac{1}{2017^{2015}}< \frac{1}{2016^{2015}};...;\frac{1}{2017}< \frac{1}{2016}\)

=> \(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}< \frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\)

=> A>B ( vì số bị chia và số chia của A và B đều dương, số bị chia của cả 2 đều là 1, cái nào có số chia nhỏ hơn thì lớn hơn)

A>B

Vì 2017>2016

Nhớ k mình nha

Vì B > 1

=> \(B=\frac{2016^{2017}}{2016^{2017}-3}>\frac{2016^{2017}+2}{2016^{2017}-3+2}=\frac{2016^{2017}+2}{2016^{2017}-1}=A\)

Vậy A=B

Ta có: \(A=\frac{2017^{100}}{1+2017+2017^2+2017^3+...+2017^{100}}\)

\(\Leftrightarrow A=\frac{\left[\left(20.100\right)+16+1\right]^{100}}{1+2017+2017^2+2017^3+...+2017^{10}}\)

        \(B=\frac{2016^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)

\(\Leftrightarrow B=\frac{\left[\left(20.100+16\right)\right]^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)

Ta có hai tổng A và B mới để so sánh:

\(A=\frac{\left[\left(20.100\right)+16+1\right]^{100}}{1+2017+2017^2+2017^3+...+2017^{100}}\)

\(B=\frac{\left[\left(20.100\right)+16\right]^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)

 Tới đây đơn giản rồi. Bạn làm tiếp đi nhé! Mẹ mình bắt tắt máy không cho làm nên đành dừng lại ở đây thôi! Thông cảm :V

=.....nha các bn. k mình nha

Ta có : \(B=\frac{2015+2016+2017}{2016+2017+2018}\) \(=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

Mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

       \(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

        \(\frac{2017}{2018}>\frac{2017}{2016+2017+2016}\)

Cộng vế theo vế, ta có : 

\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)

\(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

Ta có:

\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

Cộng vế theo vế, ta có:

\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

Vậy A >  B