BT
2
K
Khách
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DN
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 5 2023
B = \(\dfrac{2021\times13+2007+2020\times2007}{2020+2020\times520+1500\times2020}\)
B = \(\dfrac{2021\times13+2007\times\left(1+2020\right)}{2020\times\left(1+520+1500\right)}\)
B = \(\dfrac{2021\times13+2007\times2021}{2020\times2021}\)
B = \(\dfrac{2021\times\left(13+2007\right)}{2021\times2020}\)
B = \(\dfrac{2021\times2020}{2021\times2020}\)
B = 1
H9
HT.Phong (9A5)
CTVHS
15 tháng 4 2023
\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
Ta có: \(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+2+...+2\right)+\left(3+3+3+...+3\right)+...+\left(2019+2019\right)+2020\)
Trong đó có: 2020 số 1, 2019 số 2, 2018 số 3,..., 2 số 2019, 1 số 2020
Vậy: \(\left(1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+3+...+3\right)+...+2020\)
\(=1\times2020+2\times2019+3\times2018+...+2020\times1\)
\(\Rightarrow A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
\(A=\dfrac{1\times2020+2\times2019+3\times2018+...+2020\times1}{1\times2020+2\times2019+3\times2018+...+2020\times1}=1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 5 2023
A = \(\dfrac{2020}{2019}\) - \(\dfrac{2019}{2018}\) + \(\dfrac{1}{2019\times2018}\)
A = \(\dfrac{2020}{2019}\) - \(\dfrac{2019}{2018}\) + ( \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\))
A = \(\dfrac{2020}{2019}\) - \(\dfrac{2019}{2018}\) + \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)
A = ( \(\dfrac{2020}{2019}\) - \(\dfrac{1}{2019}\)) - ( \(\dfrac{2019}{2018}\) - \(\dfrac{1}{2018}\))
A = \(\dfrac{2019}{2019}\) - \(\dfrac{2018}{2018}\)
A = 1 - 1
A = 0
S
23 tháng 4 2020
\(x\times2020-x=2020\times2018+2020\)
\(\Rightarrow x\times\left(2020-1\right)=2020\times\left(2018+1\right)\)
\(\Rightarrow x\times2019=2020\times2019\)
\(\Rightarrow x=2020\)
23 tháng 4 2020
x * 2020 - x = 2020 * 2018 + 2020
x * 2020 - x * 1 = 2020 * 2018 + 2020 * 1
x * ( 2020 - 1) = 2020 * (2018 + 1)
x * 2019 = 2020 * 2019
Vậy x = 2020
DT
3
18 tháng 7 2020
Trả lời:
\(A=\frac{2}{2018.2020}+\frac{2021}{2020}-\frac{2020}{2019}\)
\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-\left(1+\frac{1}{2018}\right)\)
\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-1-\frac{1}{2018}\)
\(A=0\)
18 tháng 7 2020
\(A=\frac{2}{2018}\cdot2020+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\frac{2\cdot2020-2019}{2018}+\frac{2021}{2020}\)
\(A=\frac{2021}{2018}+\frac{2021}{2020}\)
\(A=\frac{2021\cdot\left(2020+2018\right)}{2018\cdot2020}=\frac{2021\cdot4038}{2018\cdot2020}=\frac{2021\cdot2019\cdot2}{2018\cdot1010\cdot2}=\frac{2020^2-1}{2018\cdot101\cdot10}\)
\(A=\frac{4080399}{20200180}\)
19 tháng 5 2022
a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)
NM
5
SD
1
\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\\ =\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\\ =\dfrac{2020\times4040}{2020\times4040}\\ =1\)
\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\)
\(=\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\)
\(=\dfrac{2018+2022}{4040}\)
\(=\dfrac{4040}{4040}\)
\(=1\)