so sánh P và Q biết P=10^2018 +1/10^2020 +1 và Q=10^2019 +1/10^2021 +1 29/09/2021 Bởi Samantha so sánh P và Q biết P=10^2018 +1/10^2020 +1 và Q=10^2019 +1/10^2021 +1
`P=\frac{10^{2018}+1}{10^{2020}+1}⇒100P=\frac{(10^{2020}+1)+99}{10^{2020}+1}=1+\frac{99}{10^{2020}+1}` `Q=\frac{10^{2019}+1}{10^{2021}+1}⇒100Q=\frac{(10^{2021}+1)+99}{10^{2021}+1}=1+\frac{99}{10^{2021}+1}` Vì `\frac{99}{10^{2020}+1}>\frac{99}{10^{2021}+1}` `⇒1+\frac{99}{10^{2020}+1}>1+\frac{99}{10^{2021}+1}` `⇒100P>100Q` `⇒P>Q` Vậy `P>Q`. Bình luận
#Dqnghiep P=$\frac{10^{2018}+1}{10^{2020}+1}$ ⇒ 10²P = $\frac{10².(10^{2018}+1)}{10^{2020}+1}$ = $\frac{(10^{2020}+1)+99}{10^{2020}+1}$ =$\frac{10^{2020}+1}{10^{2020}+1}$+$\frac{99}{10^{2020}+1}$ =1+$\frac{99}{10^{2020}+1}$ Q=$\frac{10^{2019}+1}{10^{2021}+1}$ ⇒ 10²P = $\frac{10²(10^{2021}+1)}{10^{2021}+1}$ =$\frac{(10^{2021}+1)+99}{10^{2021}+1}$ =$\frac{10^{2021}+1}{10^{2021}+1}$+$\frac{99}{10^{2021}+1}$ =1+$\frac{99}{10^{2021}+1}$ ⇒ 10²P > 10²Q vì $\frac{99}{10^{2020}+1}$ < $\frac{99}{10^{2021}+1}$ ⇒P > Q Bình luận
`P=\frac{10^{2018}+1}{10^{2020}+1}⇒100P=\frac{(10^{2020}+1)+99}{10^{2020}+1}=1+\frac{99}{10^{2020}+1}`
`Q=\frac{10^{2019}+1}{10^{2021}+1}⇒100Q=\frac{(10^{2021}+1)+99}{10^{2021}+1}=1+\frac{99}{10^{2021}+1}`
Vì `\frac{99}{10^{2020}+1}>\frac{99}{10^{2021}+1}`
`⇒1+\frac{99}{10^{2020}+1}>1+\frac{99}{10^{2021}+1}`
`⇒100P>100Q`
`⇒P>Q`
Vậy `P>Q`.
#Dqnghiep
P=$\frac{10^{2018}+1}{10^{2020}+1}$ ⇒ 10²P = $\frac{10².(10^{2018}+1)}{10^{2020}+1}$
= $\frac{(10^{2020}+1)+99}{10^{2020}+1}$
=$\frac{10^{2020}+1}{10^{2020}+1}$+$\frac{99}{10^{2020}+1}$
=1+$\frac{99}{10^{2020}+1}$
Q=$\frac{10^{2019}+1}{10^{2021}+1}$ ⇒ 10²P = $\frac{10²(10^{2021}+1)}{10^{2021}+1}$
=$\frac{(10^{2021}+1)+99}{10^{2021}+1}$
=$\frac{10^{2021}+1}{10^{2021}+1}$+$\frac{99}{10^{2021}+1}$
=1+$\frac{99}{10^{2021}+1}$
⇒ 10²P > 10²Q vì $\frac{99}{10^{2020}+1}$ < $\frac{99}{10^{2021}+1}$
⇒P > Q