so sánh 15^2011-1/15^2012-1 và 15^2010+1/15^2011+1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(M=-\dfrac{7}{10^{2011}}+\dfrac{-15}{10^{2012}}\) và \(N=\dfrac{-15}{10^{2011}}+\dfrac{-8}{10^{2012}}\)
Xét \(M=-\dfrac{7}{10^{2011}}-\dfrac{15}{10^{2012}}=-\dfrac{1}{10^{2011}}\left(7+\dfrac{15}{10}\right)=-\dfrac{1}{10^{2011}}\cdot\dfrac{17}{2}\).
Xét \(N=-\dfrac{15}{10^{2011}}-\dfrac{8}{10^{2012}}=-\dfrac{1}{10^{2011}}\left(15+\dfrac{8}{10}\right)=-\dfrac{1}{10^{2011}}\cdot\dfrac{79}{5}\).
Ta cũng có : \(\dfrac{M}{N}=\dfrac{-\dfrac{1}{10^{2011}}\cdot\dfrac{17}{2}}{-\dfrac{1}{10^{2011}}\cdot\dfrac{79}{5}}=\dfrac{\dfrac{17}{2}}{\dfrac{79}{5}}=\dfrac{85}{158}\)
\(\Rightarrow M=\dfrac{85}{158}N\). Mà \(\dfrac{85}{158}< 1\) nên \(M< N\).
Vậy : \(M< N\).
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2011}+\frac{2012}{2010}}\)
\(=\frac{1}{2010\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)}+\frac{1}{2011\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2012}\right)}+\frac{1}{2012\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)}\)
\(=\frac{\frac{1}{2010}}{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}+\frac{\frac{1}{2011}}{\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2012}}+\frac{\frac{1}{2012}}{\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}}\)
\(=\frac{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}=1\)
Mà \(\frac{2016}{2017}< 1\)
Vậy \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2010}+\frac{2012}{2011}}>\frac{2016}{2017}\)
dấu cần điền là : >
Vì kết quả của phép tính vế thứ 1 là 1
và phân số 2016/2017 bé hơn 1 nên ta điền dấu lớn
![](https://rs.olm.vn/images/avt/0.png?1311)
15 ^2011 -1 /15 ^2012 -1 > 15 ^2010 +1 /15 ^2011 +1
Đặt A = \(\frac{15^{2011}-1}{15^{2012}-1}\) và \(B=\frac{15^{2010}+1}{15^{2011}+1}\)
\(\Rightarrow15A=\frac{15^{2012}-15}{15^{2012}-1}=\frac{15^{2012}-1-14}{15^{2012}-1}=1-\frac{14}{15^{2012}-1}\)
\(\Rightarrow15B=\frac{15^{2011}+15}{15^{2011}+1}=\frac{15^{2011}+1+14}{15^{2011}+1}=1+\frac{14}{15^{2011}+1}\)
Ta thấy : \(1-\frac{14}{15^{2012}-1}< 1+\frac{14}{15^{2011}+1}\)
hay 15 A < 15B
=> A<B
hay \(\frac{15^{2011}-1}{15^{2012}-1}\) <\(\frac{15^{2010}+1}{15^{2011}+1}\)