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1.ĐK: n khác 2
Để A nguyên thì \(\dfrac{9}{n-2}\)nguyên <=> 9 chia hết cho n-2 hay n-2 là Ư(9) và n là số tự nhiên
Mà Ư(9)={-9;-3;-1;1;3;9}
Ta có bảng sau:
| n-2 | -9 | -3 | -1 | 1 | 3 | 9 |
| n | -7(L) | -1(L) | 1(TM) | 3(TM) | 5(TM) | 11(TM) |
Vậy n={1;3;5;9} thì A nguyên.
2.Ta xét tích:
(102016+2)(102016-3)
=104032-102016-6
(102016-1)102016
=104032-102016
104032-102016-6<104032-102016
=>(102016+2)(102016-3)<(102016-1)102016
Chia cả 2 vế cho (102016-1)(102016-3)
=>\(\dfrac{10^{2016}+2}{10^{2016}-1}< \dfrac{10^{2016}}{10^{2016}-3}\)
=>A<B
A<B bạn à . Mình chỉ phán đoán thui chứ chi tiết mình chịu . Hề Hề
Phân tích từ B ra, ta có:
B=\(\dfrac{2015+2016}{2016+2017}\)=\(\dfrac{2015}{2016+2017}\)+\(\dfrac{2016}{2016+2017}\)
Vì \(\dfrac{2015}{2016+2017}\)<\(\dfrac{2015}{2016}\) ; \(\dfrac{2016}{2016+2017}\) < \(\dfrac{2016}{2017}\)
=> B < A
a, Ta có: \(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)
\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)
\(\Rightarrow A=\dfrac{2016+2017}{2017+2018}< B=\dfrac{2016}{2017}+\dfrac{2017}{2018}\)
Vậy A < B
b, Ta có: \(\dfrac{2017}{2016+2017}< \dfrac{2017}{2016}\)
\(\dfrac{2018}{2016+2017}< \dfrac{2018}{2017}\)
\(\Rightarrow M=\dfrac{2017+2018}{2016+2017}< N=\dfrac{2017}{2016}+\dfrac{2018}{2017}\)
Vậy M < N
1. Ta có: \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(m\in Z\right)\)
\(B=\dfrac{2016^{2016}}{2016^{2016}-3}>\dfrac{2016^{2016}+2}{2016^{2016}-3+2}=\dfrac{2016^{2016}+2}{2016^{2016}-1}=A\)
Vậy A > B
2. \(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+...+\dfrac{1}{1.2}\)
= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
= \(1-\dfrac{1}{2016}\)
=\(\dfrac{2015}{2016}\)
\(A=\frac{2014^{2015}+2}{2014^{2016}+9}\)
\(2014A=\frac{2014\left(2014^{2015}+2\right)}{2014^{2016}+9}=\frac{2014^{2016}+4028}{2014^{2016}+9}=\frac{\left(2014^{2016}+9\right)+4019}{2014^{2016}+9}=\frac{2014^{2016}+9}{2014^{2016}+9}+\frac{4019}{2014^{2016}+9}=1+\frac{4019}{2014^{2016}+9}\)
\(B=\frac{2014^{2016}+2}{2014^{2017}+9}\)
\(2014B=\frac{2014\left(2014^{2016}+2\right)}{2014^{2017}+9}=\frac{2014^{2017}+4028}{2014^{2017}+9}=\frac{2014^{2017}+9+4019}{2014^{2017}+9}=\frac{2014^{2017}+9}{2014^{2017}+9}+\frac{4019}{2014^{2017}+9}=1+\frac{4019}{2014^{2017}+9}\)
Ta thấy:
\(2014^{2016}+9< 2014^{2017}+9\)
\(\Rightarrow\frac{4019}{2014^{2016}+9}>\frac{4019}{2014^{2017}+9}\)
\(\Rightarrow1+\frac{4019}{2014^{2016}+9}>1+\frac{4019}{2014^{2017}+9}\)
\(\Rightarrow A>B\)
Vậy ....
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
Ta có :
\(A=\dfrac{2016^9+3}{2016^9-1}=\dfrac{2016^9-1+4}{2016^9-1}=\dfrac{2016^9-1}{2016^9-1}+\dfrac{4}{2016^9-1}=1+\dfrac{4}{2016^9-1}\)
\(B=\dfrac{2016^9}{2016^9-4}=\dfrac{2016^9-4+4}{2016^9-4}=\dfrac{2016^9-4}{2016^9-4}+\dfrac{4}{2016^9-4}=1+\dfrac{4}{2016^9-4}\)
Vì \(1+\dfrac{4}{2016^9-1}< 1+\dfrac{4}{2016^9-4}\Rightarrow A< B\)