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DT
0
20 tháng 4 2020
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
<=> \(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
<=> \(\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
<=> x + 2015 = 0 ( vì \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x = - 2015
Vậy x = -2015.
20 tháng 4 2020
Giải phương trình :
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
\(\Rightarrow\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Mà \(\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)>0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)
3 tháng 3 2020
Đề: \(\frac{x-2}{2020}+\frac{x-3}{2019}=\frac{x-4}{2018}+\frac{x-5}{2017}\)
⇔ \(\left(\frac{x-2}{2020}-1\right)+\left(\frac{x-3}{2019}-1\right)=\left(\frac{x-4}{2018}-1\right)+\left(\frac{x-5}{2017}-1\right)\)
⇔ \(\frac{x-2022}{2020}+\frac{x-2022}{2019}=\frac{x-2022}{2018}+\frac{x-2022}{2017}\)
⇔\(\frac{x-2022}{2020}+\frac{x-2022}{2019}-\frac{x-2022}{2018}-\frac{x-2022}{2017}=0\)
⇔ \(\left(x-2022\right)\)\(\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)\) = 0
Nên x - 2022 = 0 ⇔ x = 2022
Mà \(\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)\)≠0
Vậy nghiệm của pt là x = 2022
HN
19 tháng 3 2021
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}+\dfrac{x+3}{2018}+\dfrac{x+4}{2017}+4=0\)
⇔ \(\dfrac{x+1}{2020}+1+\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1+\dfrac{x+4}{2017}+1=0\)
\(\Leftrightarrow\) \(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}=0\)
⇔ \(\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)=0\)
\(Do\) \(\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)\ne0\)
⇒ \(x+2021=0\)
⇔ \(x=-2021\)
\(Vậy\) \(x=-2021\)
12 tháng 10 2020
1. 20182 - 2017.2019 = 20182 - ( 2018 - 1 )( 2018 + 1 ) = 20182 - ( 20182 - 12 ) = 20182 - 20182 + 12 = 1
2. 20012 = ( 2000 + 1 )2 = 20002 + 2.2000.1 + 12 = 4 000 000 + 4000 + 1 = 4 004 001
3. 19992 = ( 2000 - 1 )2 = 20002 - 2.2000.1 + 12 = 4 000 000 - 4000 + 1 = 3 996 001
4. 113 = ( 10 + 1 )3 = 103 + 3.102.1 + 3.10.12 + 13 = 1000 + 300 + 30 + 1 = 1331
5. 193 = ( 20 - 1 )3 = 203 - 3.202.1 + 3.20.1 - 13 = 8000 - 1200 + 60 - 1 = 6859
6. 57.63 = ( 60 - 3 )( 60 + 3 ) = 602 - 32 = 3600 - 9 = 3591
7. 953 + 3.952.5 + 3.95.52 + 53 = ( 95 + 5 )3 = 1003 = 1 000 000
NQ
0
DB
4
8 tháng 10 2019
a,
\(2018^2-2017\cdot2019\\ =2018^2-\left(2018-1\right)\left(2018+1\right)\\ =2018^2-2018^2+1\\ =1\)
b, Đề khó nhìn bạn ạ, gõ Latex đi bạn! :)
= ( 2 + 22 + 23) + ( 24 + 25 + 26) + ............ + (22017 + 22018 + 22019) \(⋮\)7
= 2.( 1 + 2 + 22 ) + 24 (1 + 2 + 22) +..................+ 22017.( 1 + 2 + 22) \(⋮\)7
=( 2. 7 + 24.7 +.............+ 22017.7 ) \(⋮\)7
= 7 . ( 2 + 24 + ..... + 22017) \(⋮\)7
Vâỵ : ....................................................... \(⋮\)7