Tính M=\(\frac{2016^{10}+2016^{11}}{2016^{10}-2016^{11}}\)
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TN
1
4 tháng 3 2018
M=\(\frac{2016^{10}+2016^{11}}{2016^{10}-2016^{11}}=-\frac{2017}{2015}\)
NK
0
BT
1
18 tháng 10 2020
Theo đề ta có:
\(\frac{2016^{10}+2016^{11}}{2016^{10}-2016^{11}}\)
\(=\frac{2016^{10}\cdot\left(2016^0+2016^1\right)}{2016^{10}\cdot\left(2016^0-2016^1\right)}\)
\(=\frac{2016^0+2016}{2016^0-2016}\)
\(=\frac{1+2016}{1-2016}\)\(=\frac{2017}{-2015}\)\(=\frac{-2017}{2015}\)
GL
1
29 tháng 12 2019
Ta có : \(A=\frac{10^{2016}-1}{10^{2017}-11}\)
\(\Leftrightarrow10.A=\frac{10.\left(10^{2016}-1\right)}{10^{2017}-11}=\frac{10^{2017}-10}{10^{2017}-11}\)
\(=\frac{10^{2017}-11+1}{10^{2017}-11}=1+\frac{1}{10^{2017}-11}\)
Lại có : \(B=\frac{10^{2016}+1}{10^{2017}+9}\)
\(\Leftrightarrow10.B=\frac{10\left(10^{2016}+1\right)}{10^{2017}+9}=\frac{10^{2017}+10}{10^{2017}+9}\)
\(=\frac{10^{2017}+9+1}{10^{2017}+9}=1+\frac{1}{10^{2017}+9}\)
Do : \(10^{2017}-11< 10^{2017}+9\) \(\Rightarrow\frac{1}{10^{2017}-11}>\frac{1}{10^{2017}+9}\)
\(\Rightarrow1+\frac{1}{10^{2017}-11}>1+\frac{1}{10^{2017}+9}\)
hay \(A>B\)
Vậy : \(A>B\)
PQ
2
23 tháng 12 2017
Ta có:2016A=\(\frac{2016\left(2016^{10}+1\right)}{2016^{11}+1}=\frac{2016^{11}+2016}{2016^{11}+1}=\frac{2016^{11}+1+2015}{2016^{11}+1}=1+\frac{2015}{2016^{11}+1}\)
2016B=\(\frac{2016\left(2016^{11}+1\right)}{2016^{12}+1}=\frac{2016^{12}+2016}{2016^{12}+1}=\frac{2016^{12}+1+2015}{2016^{12}+1}=1+\frac{2015}{2016^{12}+1}\)
Vì \(2016^{12}+1>2016^{11}+1\) nên \(\frac{2015}{2016^{12}+1}< \frac{2015}{2016^{11}+1}\)
\(\Rightarrow1+\frac{2015}{2016^{12}+1}< 1+\frac{2015}{2016^{11}+1}\)\(\Rightarrow2016B< 2016A\Rightarrow B< A\)
23 tháng 12 2017
Ta có :
\(B=\frac{2016^{11}+1}{2016^{12}+1}>\frac{2016^{11}+1+9}{2016^{12}+1+9}=\frac{2016^{11}+10}{2016^{12}+10}=\frac{10\left(2016^{10}+1\right)}{10\left(2016^{11}+1\right)}=\frac{2016^{10}+1}{2016^{11}+1}=A\)
Vậy \(A< B\)
4 tháng 2 2017
Ta có \(S=\frac{2016}{5}+\frac{2016}{10}+\frac{2016}{30}+...+\frac{2016}{47530}+\frac{2016}{48150}\)
\(=\frac{2016}{5}.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9506}+\frac{1}{9630}\right)\)
\(=\frac{2016}{5}.\left(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
\(=\frac{2016}{5}.\left(\frac{1}{1}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{2016}{5}.\left(1+1-\frac{1}{99}\right)\)
\(=\frac{2016}{5}.\frac{197}{99}\)
\(=\frac{44128}{55}\)
\(M=\frac{2016^{10}+2016^{11}}{2016^{10}-2016^{11}}=\frac{2016^{10}\left(1+2016\right)}{2016^{10}\left(1-2016\right)}=\frac{-2017}{2015}\)