So sánh A= $\dfrac{7^2^0^1^7+1}{7^2018+1} và B=$\dfrac{7^2^0^1^8 +1}{7^2019+1} Lưu ý : cộng 1 ko phải số mũ nhé !!

Đáp án:

 $A>B$

Giải thích các bước giải:

$A=\dfrac{7^{2017}+1}{7^{2018}+1}\\
\Rightarrow 7A=\dfrac{7.(7^{2017}+1)}{7^{2018}+1}\\
=\dfrac{7^{2018}+7}{7^{2018}+1}\\
=\dfrac{7^{2018}+1+6}{7^{2018}+1}\\
=\dfrac{7^{2018}+1}{7^{2018}+1}+\dfrac{6}{7^{2018}+1}\\
=1+\dfrac{6}{7^{2018}+1}\\
B=\dfrac{7^{2018}+1}{7^{2019}+1}\\
\Rightarrow 7B=\dfrac{7.(7^{2018}+1)}{7^{2019}+1}\\
=\dfrac{7^{2019}+7}{7^{2019}+1}\\
=\dfrac{7^{2019}+1+6}{7^{2019}+1}\\
=\dfrac{7^{2019}+1}{7^{2019}+1}+\dfrac{6}{7^{2019}+1}\\
=1+\dfrac{6}{7^{2019}+1}$
Vì $7^{2018}+1<7^{2019}+1$
$\Rightarrow \dfrac{1}{7^{2018}+1}>\dfrac{1}{7^{2019}+1}\\ \Rightarrow \dfrac{6}{7^{2018}+1}>\dfrac{6}{7^{2019}+1}\\
\Rightarrow 1+\dfrac{6}{7^{2018}+1}>1+\dfrac{6}{7^{2019}+1}\\
\Rightarrow 7A>7B\\
\Rightarrow A>B$