Ta có :
\(M=\dfrac{2018^{2017}+1}{2018^{2018}+1}< 1\)
\(\Rightarrow M< \dfrac{2018^{2017}+1+2017}{2017^{2018}+1+2017}=\dfrac{2018^{2017}+2018}{2017^{2018}+2018}=\dfrac{2018\left(2018^{2016}+1\right)}{2018\left(2018^{2017}+1\right)}=\dfrac{2018^{2016}+1}{2018^{2017}+1}=N\)
\(\Rightarrow M< N\)
Giải:
Ta có:
\(2018M=\dfrac{\left(2018^{2017}+1\right)2018}{2018^{2018}+1}.\)
\(2018M=\dfrac{2018^{2018}+2018}{2018^{2018}+1}.\)
\(2018M=\dfrac{\left(2018^{2018}+1\right)+2017}{2018^{2018}+1}.\)
\(2018M=\dfrac{2018^{2018}+1}{2018^{2018}+1}+\dfrac{2017}{2018^{2018}+1}.\)
\(2018M=1+\dfrac{2017}{2018^{2018}+1}._{\left(1\right)}\)
Ta lại có:
\(2018N=\dfrac{\left(2018^{2016}+1\right)2018}{2018^{2017}+1}.\)
\(2018N=\dfrac{2018^{2017}+2018}{2018^{2017}+1}.\)
\(2018N=\dfrac{\left(2018^{2017}+1\right)+2017}{2018^{2017}+1}.\)
\(2018N=\dfrac{2018^{2017}+1}{2018^{2017}+1}+\dfrac{2017}{2018^{2017}+1}.\)
\(2018N=1+\dfrac{2017}{2018^{2017}+1}._{\left(2\right)}\)
Và \(\dfrac{2017}{2018^{2018}+1}< \dfrac{2017}{2018^{2017}+1}._{\left(3\right)}\)
Từ \(_{\left(1\right);\left(2\right)}\) và \(_{\left(3\right)}\Rightarrow2018M< 2018N\Rightarrow M< N.\)
Vậy......
~ Học tốt!!! ~
