A = \(\dfrac{2006^{2016}+1}{2006^{2017}+1}\) 

A \(\times\) 2006 = \(\dfrac{(2006^{2016}+1)\times2006}{2006^{2017}+1}\)

A \(\times\) 2006 = \(\dfrac{2006^{2017}+2006}{2006^{2017}+1}\)

A \(\times\)2006 = 1 + \(\dfrac{2006}{2006^{2017}+1}\)

B = \(\dfrac{2006^{2015}+1}{2006^{2016}+1}\)

B \(\times\) 2006 =  \(\dfrac{\left(2006^{2015}+1\right)\times2006}{2006^{2016}+1}\)

B \(\times\) 2006 = \(\dfrac{2006^{2016}+2006}{2006^{2016}}\)

B \(\times\) 2006 = 1 + \(\dfrac{2006}{2006^{2016}+1}\)

Vì \(\dfrac{2006}{2006^{2016}+1}\) > \(\dfrac{2006}{2006^{2017}+1}\)

=> B \(\times\) 2006 > A \(\times\) 2006

B > A 

khó wa

 Ko bt

 A=2016/2017+2017/2018

 Do 2016/2017<1,2017/2018<1=> A<2 Hay A<B

Câu b tương tự ha

Cảm ơn !

Ta có : \(A=\left(\left(-2015\right)^{2016}.-2016^{2017}+\left(-2016\right)^{2017}.-2015^{2016}\right).\left(-2017\right)^{2018}\)

\(=\left(2015^{2016}.-2016^{2017}-2016^{2017}.-2015^{2016}\right).2017^{2018}\)

\(=\left(2015^{2016}-2015^{2016}\right).2017^{2018}.\left(-2016^{2017}\right)\)

\(=0.2017^{2018}.\left(-2016^{2017}\right)=0\)

Giải:

\(A=\left[\left(-2015\right)^{2016}.\left(-2016^{2017}\right)+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\) 

\(A=\left[2015^{2016}.\left(-2016\right)^{2017}+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\) 

\(A=\left[2015^{2016}+\left(-2015^{2016}\right)\right].\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\) 

\(A=0.\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\) 

\(A=0\)

Bằng \(-\frac{2}{11}\)

a)\(=\frac{2017}{2016}.\frac{3}{4}-\frac{1}{2016}.\frac{3}{4}\)

\(=\frac{3}{4}\left(\frac{2017}{2016}-\frac{1}{2016}\right)\)

\(=\frac{3}{4}.1\)

\(=\frac{3}{4}\)

b)\(=\frac{2015}{2016}\left(\frac{1}{2}+\frac{1}{3}-\frac{5}{6}\right)\)

\(=\frac{2015}{2016}.0\)

\(=0\)