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a/ \(A=2018\cdot2018\)
\(=\left(2019-1\right)\cdot2018=2019\cdot2018-2018\)
\(B=2017\cdot2019\)
\(=\left(2018-1\right)\cdot2019=2018\cdot2019-2019\)
\(\Rightarrow A>B\)
b/
\(A=2018\cdot2019\)
\(=\left(2017+1\right)\cdot2019=2017\cdot2019+2019\)
\(B=2017\cdot2020\)
\(=2017\cdot\left(2019+1\right)=2017\cdot2019+2017\)
\(\Rightarrow A>B\)
26 x 84 + 74 x 85
= 26 x 84 + 74 x 84 + 74
= 84 x (26 + 74) + 74
= 84 x 100 + 74
= 840 + 74
= 914
\(\frac{2017\times2018+2019}{2019\times2018-2017}\)
= \(\frac{2019\times2018-2\times2018+2019}{2019\times2018-2017}\)
= \(\frac{2019\times2018-4036+2019}{2019\times2018-2017}\)
= \(\frac{2019\times2018-2017}{2019\times2018-2017}\)
= 1
\(A=1-3+5-7+......-2019+2021-2023\)
\(A=\left(1-3\right)+\left(5-7\right)+....+\left(2021-2023\right)\)
\(A=-2+\left(-2\right)+....+\left(-2\right)\left(506 cặp\right)\)
\(A=-2.506\)
\(A=-1012\)
*) A=(1-3)+(5-7)+....+(2021-2023)
<=> A=-2+(-2)+...+(-2)
Dãy A có (2023-1):2+1=1012 số số hạng
=> Có 506 số (-2)
=> A=(-2).506=-1012
bạn nào làm được thì giúp mình với còn bài này thì mình không biết làm. sorry nha
AI NÓI TỚ NÓI SAI, CÓ NÓI VỀ BÀI ĐÂU MÀ SAI ĐIÊN À MẤY BẠN KIA
Đáp án: 1
TA CÓ:
E=1+(2-3-4+5)+(6-7-8+9)+.......+(2018-2019-2020+2021)
E=1+0+0+0+.....+0
E=1
K CHO MIK NHAAAAA
Ta có: \(B=\dfrac{2017+2018+2019}{2018+2019+2020}=\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2019+2020}\)
Mà \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019+2020}\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019+2020}\)
\(\dfrac{2019}{2020}>\dfrac{2019}{2018+2019+2020}\)
\(\Rightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}>\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2919+2020}\)
\(\Rightarrow A>B.\)
Vậy \(A>B.\)
a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)
= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)
= 1
@Nguyen Thi Ngoc Linh