Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
A=2016/2017+2017/2018
Do 2016/2017<1,2017/2018<1=> A<2 Hay A<B
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:
\(A=\left[\left(-2015\right)^{2016}.\left(-2016^{2017}\right)+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\)
\(A=\left[2015^{2016}.\left(-2016\right)^{2017}+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\)
\(A=\left[2015^{2016}+\left(-2015^{2016}\right)\right].\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\)
\(A=0.\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\)
\(A=0\)
=2015-(2015-2016)-2016+22017-2015-22015/22014-(1-4)-3-(5+6)+11
=(2015-2015)+(2016-2016)+22-2+3-3-11+11
=0+0+(4-2)+(3-3)-(11-11)
=2
Vì:
khi tính bài toán 2015/2016 + 2016/2017 + 2017/2018 + 2018/2019 + 2019/2020 + 2020/2015 này ra thì ta được con số là 6,000003688 con số này phải lớn hơn số 6 nên: 6,000003688 > 6
Vì:khi tính bài toán 2015/2016+2016/2017+2017/2018+2018/2019+ 2019/2020+2020/2015 ta ra được là: 6,000003688 nên: 6,000003688 > 6
A= | -1| - 2009 + |-4014|
A= 1-2009+4014
A= 2006