Doris 837 Questions 2k Answers 0 Best Answers 19 Points View Profile0 Doris Asked: Tháng Mười 26, 20202020-10-26T10:44:22+00:00 2020-10-26T10:44:22+00:00In: Môn ToánSo sánh: 2020^2019 + 2020^2020 và 2021^2020.0 So sánh: 2020^2019 + 2020^2020 và 2021^2020. ShareFacebookRelated Questions Một hình thang có đáy lớn là 52cm ; đáy bé kém đáy lớn 16cm ; chiều cao kém đáy ... Useful news and important articles APROTININ FROM BOVINE LUNG CELL CULTURE купить онлайн2 AnswersOldestVotedRecentKhánh Gia 852 Questions 2k Answers 0 Best Answers 19 Points View Profile Giakhanh 2020-10-26T10:45:44+00:00Added an answer on Tháng Mười 26, 2020 at 10:45 sáng `2020^2019+2020^2020=2020^2019 . (1+2020)=2021 . 2020^2019``2021^2020=2021 . 2021^2019`Do `0<2020<2021``=> 2021 . 2020^2019<2021 . 2021^2019``=> 2020^2019+2020^2020<2021^2020`0Reply Share ShareShare on FacebookCalantha 883 Questions 2k Answers 1 Best Answer 26 Points View Profile Calantha 2020-10-26T10:46:16+00:00Added an answer on Tháng Mười 26, 2020 at 10:46 sáng Đáp án:$2021^{2020} > 2020^{2019} + 2020^{2020}$Giải thích các bước giải:$+) \quad 2020^{2019} + 2020^{2020}$$= 2020^{2019} + 2020^{2019}.2020$$= 2020^{2019}.2021$$+) \quad 2021^{2020}$$= 2021^{2019}.2021$Do $2021 > 2020$$\to 2021^{2019} > 2020^{2019}$$\to 2021^{2019}.2021 > 2020^{2019}.2021$$\to 2021^{2020} > 2020^{2019} + 2020^{2020}$0Reply Share ShareShare on FacebookLeave an answerLeave an answerHủy By answering, you agree to the Terms of Service and Privacy Policy .* Lưu tên của tôi, email, và trang web trong trình duyệt này cho lần bình luận kế tiếp của tôi.
Khánh Gia
`2020^2019+2020^2020=2020^2019 . (1+2020)=2021 . 2020^2019`
`2021^2020=2021 . 2021^2019`
Do `0<2020<2021`
`=> 2021 . 2020^2019<2021 . 2021^2019`
`=> 2020^2019+2020^2020<2021^2020`
Calantha
Đáp án:
$2021^{2020} > 2020^{2019} + 2020^{2020}$
Giải thích các bước giải:
$+) \quad 2020^{2019} + 2020^{2020}$
$= 2020^{2019} + 2020^{2019}.2020$
$= 2020^{2019}.2021$
$+) \quad 2021^{2020}$
$= 2021^{2019}.2021$
Do $2021 > 2020$
$\to 2021^{2019} > 2020^{2019}$
$\to 2021^{2019}.2021 > 2020^{2019}.2021$
$\to 2021^{2020} > 2020^{2019} + 2020^{2020}$