$So$ $sánh$ $\frac{2018^{2019}-2}{2018^{2020}-2}$ $và$ $\frac{2018^{2019}-2+2020}{2018^{2020}-2+2020}$ 15/10/2021 Bởi Kennedy $So$ $sánh$ $\frac{2018^{2019}-2}{2018^{2020}-2}$ $và$ $\frac{2018^{2019}-2+2020}{2018^{2020}-2+2020}$
Đáp án: $A<B$ Giải thích các bước giải: Đặt $A=\dfrac{2018^{2019}-2}{2018^{2020}-2}\\\Rightarrow 2018A=\dfrac{2018^{2020}-2.2018}{2018^{2020}-2}\\=\dfrac{2018^{2020}-2-4034}{2018^{2020}-2}=1-\dfrac{4034}{2018^{2020}-2}\\B=\dfrac{2018^{2019}-2+2020}{2018^{2020}-2+2020}\\=\dfrac{2018^{2019}+2018}{2018^{2020}+2018}\\=\dfrac{2018^{2018}+1}{2018^{2019}+1}\\\Rightarrow 2018B=\dfrac{2018^{2019}+2018}{2018^{2019}+1}\\=\dfrac{2018^{2019}+1+2017}{2018^{2019}+1}\\=1+\dfrac{2017}{2018^{2019}+1}$Vì $1-\dfrac{4034}{2018^{2020}-2}<1<1+\dfrac{2017}{2018^{2019}+1}$Nên $2018A<2018B\Rightarrow A<B$ Bình luận
Đáp án:
$A<B$
Giải thích các bước giải:
Đặt
$A=\dfrac{2018^{2019}-2}{2018^{2020}-2}\\
\Rightarrow 2018A=\dfrac{2018^{2020}-2.2018}{2018^{2020}-2}\\
=\dfrac{2018^{2020}-2-4034}{2018^{2020}-2}=1-\dfrac{4034}{2018^{2020}-2}\\
B=\dfrac{2018^{2019}-2+2020}{2018^{2020}-2+2020}\\
=\dfrac{2018^{2019}+2018}{2018^{2020}+2018}\\
=\dfrac{2018^{2018}+1}{2018^{2019}+1}\\
\Rightarrow 2018B=\dfrac{2018^{2019}+2018}{2018^{2019}+1}\\
=\dfrac{2018^{2019}+1+2017}{2018^{2019}+1}\\
=1+\dfrac{2017}{2018^{2019}+1}$
Vì $1-\dfrac{4034}{2018^{2020}-2}<1<1+\dfrac{2017}{2018^{2019}+1}$
Nên $2018A<2018B\Rightarrow A<B$