Đáp án:
$M>N$
Giải thích các bước giải:
$a)\dfrac{1}{5}M=\dfrac{5^{2018}+1}{5^{2018}+5}=\dfrac{5^{2018}+5-4}{5^{2018}+5}\\
\Rightarrow \dfrac{1}{5}M=1-\dfrac{4}{5^{2018}+5}\\
\dfrac{1}{5}N=\dfrac{5^{2017}+1}{5^{2017}+5}=\dfrac{5^{2017}+5-4}{5^{2017}+5}\\
\Rightarrow \dfrac{1}{5}N=1-\dfrac{4}{5^{2017}+5}$
Ta có $5^{2018}+5>5^{2017}+5$
$\Rightarrow \dfrac{4}{5^{2018}+5}<\dfrac{4}{5^{2017}+5}\\
\Rightarrow 1-\dfrac{4}{5^{2018}+5}>1-\dfrac{4}{5^{2017}+5}\\
\Rightarrow \dfrac{1}{5}M>\dfrac{1}{5}N\\
\Rightarrow M>N$
