So sánh: A=$\frac{20^10+1}{20^10-1}$ và B=$\frac{20^10-1}{20^10-3}$ 15/10/2021 Bởi Raelynn So sánh: A=$\frac{20^10+1}{20^10-1}$ và B=$\frac{20^10-1}{20^10-3}$
Đáp án: $ A<B$ Giải thích các bước giải: $A=\dfrac{20^{10}+1}{20^{10}-1}\\=\dfrac{20^{10}-1+2}{20^{10}-1}\\=1+\dfrac{2}{20^{10}-1}B=\dfrac{20^{10}-1}{20^{10}-3}\\=\dfrac{20^{10}-3+2}{20^{10}-3}\\=1+\dfrac{2}{20^{10}-3}$Do $20^{10}-1>20^{10}-3$$\Rightarrow \dfrac{1}{20^{10}-1}<\dfrac{1}{20^{10}-3}\\\Rightarrow \dfrac{2}{20^{10}-1}<\dfrac{2}{20^{10}-3}\\\Rightarrow 1+\dfrac{2}{20^{10}-1}<1+\dfrac{2}{20^{10}-3}\\\Rightarrow A<B$ Bình luận
Đáp án:
$ A<B$
Giải thích các bước giải:
$A=\dfrac{20^{10}+1}{20^{10}-1}\\
=\dfrac{20^{10}-1+2}{20^{10}-1}\\
=1+\dfrac{2}{20^{10}-1}
B=\dfrac{20^{10}-1}{20^{10}-3}\\
=\dfrac{20^{10}-3+2}{20^{10}-3}\\
=1+\dfrac{2}{20^{10}-3}$
Do $20^{10}-1>20^{10}-3$
$\Rightarrow \dfrac{1}{20^{10}-1}<\dfrac{1}{20^{10}-3}\\
\Rightarrow \dfrac{2}{20^{10}-1}<\dfrac{2}{20^{10}-3}\\
\Rightarrow 1+\dfrac{2}{20^{10}-1}<1+\dfrac{2}{20^{10}-3}\\
\Rightarrow A<B$