so sánh 10^2020+1/10^2021+1 và 10^2020-11/10^2021-10 16/07/2021 Bởi Iris so sánh 10^2020+1/10^2021+1 và 10^2020-11/10^2021-10
A=$\frac{10^{2020}+1}{10^{2021}+1}$ 10A=$\frac{10(10^{2020}+1)}{10^{2021}+1}$ =$\frac{10^{2021}+10}{10^{2021}+1}$ =$\frac{10^{2021}+1+9}{10^{2021}+1}$ =1+$\frac{9}{10^{2021}+1}$ B=$\frac{10^{2020}-11}{10^{2021}-10}$ 10B=$\frac{10(10^{2020}-11)}{10^{2021}-10}$ =$\frac{10^{2021}-110}{10^{2021}-10}$ =$\frac{10^{2021}-10-100}{10^{2021}-10}$ =1 – $\frac{100}{10^{2021}-10}$ Ta có: $\frac{9}{10^{2021}+1}$ > $\frac{100}{10^{2021}-10}$ ⇒$\frac{9}{10^{2021}+1}$ > – $\frac{100}{10^{2021}-10}$ ⇒1+$\frac{9}{10^{2021}+1}$ > 1 – $\frac{100}{10^{2021}-10}$ ⇒ 10A> 10B ⇒A>B Bình luận
Đáp án: Giải thích các bước giải: Vì `10^2020` = `10^2020` ⇒`10^2020` + 1/`10^2021+1` < 10^2020-11/`10^2021-10` Vậy `10^2020` + 1/`10^2021+1` < 10^2020-11/`10^2021-10` Bình luận
A=$\frac{10^{2020}+1}{10^{2021}+1}$
10A=$\frac{10(10^{2020}+1)}{10^{2021}+1}$
=$\frac{10^{2021}+10}{10^{2021}+1}$
=$\frac{10^{2021}+1+9}{10^{2021}+1}$
=1+$\frac{9}{10^{2021}+1}$
B=$\frac{10^{2020}-11}{10^{2021}-10}$
10B=$\frac{10(10^{2020}-11)}{10^{2021}-10}$
=$\frac{10^{2021}-110}{10^{2021}-10}$
=$\frac{10^{2021}-10-100}{10^{2021}-10}$
=1 – $\frac{100}{10^{2021}-10}$
Ta có: $\frac{9}{10^{2021}+1}$ > $\frac{100}{10^{2021}-10}$
⇒$\frac{9}{10^{2021}+1}$ > – $\frac{100}{10^{2021}-10}$
⇒1+$\frac{9}{10^{2021}+1}$ > 1 – $\frac{100}{10^{2021}-10}$
⇒ 10A> 10B ⇒A>B
Đáp án:
Giải thích các bước giải:
Vì `10^2020` = `10^2020`
⇒`10^2020` + 1/`10^2021+1` < 10^2020-11/`10^2021-10`
Vậy `10^2020` + 1/`10^2021+1` < 10^2020-11/`10^2021-10`