So sánh 2018/2019 + 2019/2020 + 2020/2021:3
Những câu hỏi liên quan
12 tháng 2 2023 lúc 17:57
So sánh A=\(\dfrac{2018}{2019}\)+\(\dfrac{2019}{2020}\)+\(\dfrac{2020}{2021}\)+\(\dfrac{2021}{2018}\)với 4
Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$
$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$
$> 4+0+0+0+0=4$
Đúng 2
Bình luận (0)
A=2017/2018+2020/2019
B=2018/2019+2021/2020
So sánh A và B
Xét 2017 /2018 và 2018/2019
1-2017/2018=1/2018
1-2018/2019=1/2019
mà 1/2018>1/2019=>2017/2018<2018/2019
Tương tự có:2020/2019>2021/2020
=>2017/2018+2010/2019<2018/2019+2021/2020
so sánh: 2018^2019+1/2018^2020+1 và 2018^2020+1/2018^2021+1
2018^2019+1/2018^2020+1 bé hơn 2018^2020+1/2018^2021+1
Bài tập: So Sánh
M= \(\frac{2018}{2019}\)+\(\frac{2019}{2020}\)+\(\frac{2020}{2021}\)
N=\(\frac{2018+2019+2020}{2019+2020+2021}\)
Ta có :
\(N=\frac{2018+2019+2020}{2019+2020+2021}\)
\(=\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)
Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Leftrightarrow M>N\)
Đúng 0
Bình luận (0)
Trả lời:
Ta có:
\(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}>\frac{2018+2019+2020}{2019+2020+2021}\)
hay \(M>N\)
Vậy \(M>N\)
Ta có :
N = \(\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)
Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Rightarrow M>N\)
Bài 3: Không quy đồng hãy so sánh các phân số sau: a, 2019/2020 và 2021/2022 b, 2019/2017 và 2021/2019 c, 201/202 và 135/137 d, 2019/2018 và 2021/2019
so sánh P=2019/2020+2020/2021+2021/2022 và Q=2019+2020+2021/2020+2021+2022
17 tháng 4 2021 lúc 17:53
So sánh M = \(\dfrac{2019}{2020}+\dfrac{2020}{2021}\) và N = \(\dfrac{2019+2020}{2020+2021}\)
Giải:
Ta có: N=2019+2020/2020+2021
=>N=2019/2020+2021 + 2020/2020+2021
Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021
=>M>N
Vậy ...
Chúc bạn học tốt!
Đúng 1
Bình luận (0)
Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
Đúng 1
Bình luận (0)
So sánh A và B, biết:
A= 2018 x 2020 + 2021
và B= 2019 x 2019 + 2021
\(A=2018\times2020+2021\) và \(B=2019\times2019+2021\)
\(A=2018\times2019+2018+2021\)
\(B=2018\times2019+2019+2021\)
Vì \(2019>2018\Rightarrow A< B\)
Ta có :
2018 x 2020 = 2018 x ( 2019 + 1 ) = 2018 + 2018 x 2019 < 2019 + 2018 x 2019 = 2019 x ( 2018 + 1 )
= 2019 x 2019
=> 2018 x 2020 < 2019 x 2019
=> 2018 x 2020 + 2021 < 2019 x 2019 + 2021
=> A < B
\(A=2018.2020+2021=2018.2019+2018+2021\)
\(B=2019.2019+2021=2018.2019+2019+2021\)
\(\Rightarrow A< B\)
Xem thêm câu trả lời
2021/2022 + 2020/2021 + 2019/2020 + 2018/2019 + 2017/2018 rôì so sánh với 5
A = 2021/2022+2020/2021+2019/2020+2018/2019+2017/2018
A<2022/2022+2021/2021+2020/2020+2019/2019+2018/2018
A<1+1+1+1+1
A<5
Đúng 0
Bình luận (0)