A=10^2018+1/10^2019+1 B=10^2019+1/10^2020+1 12/08/2021 Bởi Clara A=10^2018+1/10^2019+1 B=10^2019+1/10^2020+1
Tham khảo Xét `A` `⇒10A=\frac{10^{2019}+10}{10^{2019}+1}` `⇒10A=1+\frac{9}{10^{2019}+1}` Xét `B` `⇒10B=\frac{10^{2020}+10}{10^{2020}+1}` `⇒10B=1+\frac{9}{10^{2020}+1}` Có `\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}` `⇒1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}` Hay `10A>10B` `⇒A>B` `\text{©CBT}` Bình luận
Đáp án: Ta có: `10A=10.(10^{2018}+1)/(10^{2019}+1)=(10^{2019}+10)/(10^{2019}+1)` `=(10^{2019}+1)/(10^{2019}+1)+9/(10^{2019}+1)=1+9/(10^{2019}+1)` `10B=10.(10^{2019}+1)/(10^{20120}+1)=(10^{2020}+10)/(10^{2020}+1)` `=(10^{2020}+1)/(10^{2019}+1)+9/(10^{2020}+1)=1+9/(10^{2020}+1)` Mà `10^{2019}<10^{2020}⇒10^{2019}+1<10^{2020}+1` `⇒9/(10^{2019}+1)+1>9/(10^{2020}+1)+1` `⇒ 10A>10B⇒A>B` Bình luận
Tham khảo
Xét `A`
`⇒10A=\frac{10^{2019}+10}{10^{2019}+1}`
`⇒10A=1+\frac{9}{10^{2019}+1}`
Xét `B`
`⇒10B=\frac{10^{2020}+10}{10^{2020}+1}`
`⇒10B=1+\frac{9}{10^{2020}+1}`
Có `\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}`
`⇒1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}`
Hay `10A>10B`
`⇒A>B`
`\text{©CBT}`
Đáp án:
Ta có:
`10A=10.(10^{2018}+1)/(10^{2019}+1)=(10^{2019}+10)/(10^{2019}+1)`
`=(10^{2019}+1)/(10^{2019}+1)+9/(10^{2019}+1)=1+9/(10^{2019}+1)`
`10B=10.(10^{2019}+1)/(10^{20120}+1)=(10^{2020}+10)/(10^{2020}+1)`
`=(10^{2020}+1)/(10^{2019}+1)+9/(10^{2020}+1)=1+9/(10^{2020}+1)`
Mà `10^{2019}<10^{2020}⇒10^{2019}+1<10^{2020}+1`
`⇒9/(10^{2019}+1)+1>9/(10^{2020}+1)+1`
`⇒ 10A>10B⇒A>B`