BÀI 1 : A = 10^2019 + 1 / 10^2020 + 1 và B = 10^2020 + 1 / 10^2021 + 1 11/08/2021 Bởi Hailey BÀI 1 : A = 10^2019 + 1 / 10^2020 + 1 và B = 10^2020 + 1 / 10^2021 + 1
Tham khảo Xét `A` `⇒10A=\frac{10^{2020}+10}{10^{2020}+1}` `⇒10A=1+\frac{9}{10^{2020}+1}` Xét `B` `⇒10B=\frac{10^{2021}+10}{10^{2021}+1}` `⇒10B=1+\frac{9}{10^{2021}+1}` Có `\frac{9}{10^{2020}+1}>\frac{9}{10^{2021}+1}` `⇒1+\frac{9}{10^{2020}+1}>1+\frac{9}{10^{2021}+1}` Hay `10A>10B` `⇒A>B` `\text{©CBT}` Bình luận
Đáp án + Giải thích các bước giải: `A = (10^2019 + 1 )/ (10^2020 + 1)=>10A=(10(10^2019+1))/(10^2020 + 1)=(10^2020+10)/(10^2020 + 1)=((10^2020+1)+9)/(10^2020 + 1)=1+9/(10^2020 + 1)` `B =(10^2020 + 1 )/( 10^2021 + 1)=>10B=(10(10^2020 + 1))/( 10^2021 + 1)=(10^2021+10)/(10^2021+1)=((10^2021+1)+9)/(10^2021+1)=1+9/(10^2021+1)` Vì `10^2020+1<10^2021+1` Nên `9/(10^2020+1)>9/(10^2021+1)` `=>1+9/(10^2020+1)>1+9/(10^2021+1)` `=>10A>10B` Vậy `A>B` Bình luận
Tham khảo
Xét `A`
`⇒10A=\frac{10^{2020}+10}{10^{2020}+1}`
`⇒10A=1+\frac{9}{10^{2020}+1}`
Xét `B`
`⇒10B=\frac{10^{2021}+10}{10^{2021}+1}`
`⇒10B=1+\frac{9}{10^{2021}+1}`
Có `\frac{9}{10^{2020}+1}>\frac{9}{10^{2021}+1}`
`⇒1+\frac{9}{10^{2020}+1}>1+\frac{9}{10^{2021}+1}`
Hay `10A>10B`
`⇒A>B`
`\text{©CBT}`
Đáp án + Giải thích các bước giải:
`A = (10^2019 + 1 )/ (10^2020 + 1)=>10A=(10(10^2019+1))/(10^2020 + 1)=(10^2020+10)/(10^2020 + 1)=((10^2020+1)+9)/(10^2020 + 1)=1+9/(10^2020 + 1)`
`B =(10^2020 + 1 )/( 10^2021 + 1)=>10B=(10(10^2020 + 1))/( 10^2021 + 1)=(10^2021+10)/(10^2021+1)=((10^2021+1)+9)/(10^2021+1)=1+9/(10^2021+1)`
Vì `10^2020+1<10^2021+1`
Nên `9/(10^2020+1)>9/(10^2021+1)`
`=>1+9/(10^2020+1)>1+9/(10^2021+1)`
`=>10A>10B`
Vậy `A>B`