Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta có :
A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)
B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)
\(\Leftrightarrow\) B < A
A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)
10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2
B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)
10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2
10A > 2 > 10B ⇒ 10A>10B ⇒ A>B
Ta có: \(A=2019.2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1\)
\(B=2020^2\)
=> A < B
Ta có \(\left(\sqrt{2018}+\sqrt{2020}\right)^2=4038+2\sqrt{4076360}\) và \(\left(2\sqrt{2019}\right)^2=8076=4038+4038\)
Mà \(\left(2\sqrt{4076360}\right)^2=16305440\) và \(4038^2=16305444\)
\(\Rightarrow2\sqrt{4076360}< 4038\)
\(\Rightarrow\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)
\(\left(\sqrt{2018}+\sqrt{2020}\right)^2=4038+2\cdot\sqrt{2018\cdot2020}\)
\(\left(2\sqrt{2019}\right)^2=8076=4038+4038\)
mà \(2\cdot\sqrt{2018\cdot2020}< 4038\)
nên \(\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)
\(A=2019\times2021=\left(2021-1\right)\times\left(2021+1\right)=2021^2-1< 2021^2=B.\)
bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
\(8^2=64=32+2\sqrt{16^2}\)
\(\left(\sqrt{15}+\sqrt{17}\right)^2=32+2\sqrt{15.17}=32+2\sqrt{\left(16-1\right)\left(16+1\right)}\)
\(=32+2\sqrt{16^2-1}\)
\(< =>8^2>\left(\sqrt{15}+\sqrt{17}\right)^2\)
\(8>\sqrt{15}+\sqrt{17}\)
\(\left(\sqrt{2019}+\sqrt{2021}\right)^2=4040+2\sqrt{2019.2021}\)
\(=4040+2\sqrt{\left(2020-1\right)\left(2020+1\right)}=4040+2\sqrt{2020^2-1}\)
\(\left(2\sqrt{2020}\right)^2=8080=4040+2\sqrt{2020^2}\)
\(< =>\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)
mik chọn điền
<
mik lười chép ại đề bài