Sửa đề: \(\dfrac{x+4}{2018}+\dfrac{x+3}{2019}+\dfrac{x+2}{2020}+\dfrac{x+1}{2021}=-4\)
=>\(\left(\dfrac{x+4}{2018}+1\right)+\left(\dfrac{x+3}{2019}+1\right)+\left(\dfrac{x+2}{2020}+1\right)+\left(\dfrac{x+1}{2021}+1\right)=-4+4=0\)
=>\(\dfrac{x+2022}{2018}+\dfrac{x+2022}{2019}+\dfrac{x+2022}{2020}+\dfrac{x+2022}{2021}=0\)
=>x+2022=0
=>x=-2022
TÌM X BIẾT:
\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)
HELP ME!!
Ai nhanh và đúng mik tick.
x+4/2018+x+3/2019=x+2/2020+x+1/2021
tìm x biết:
(x+1)/2018+(x+1)/2019+(x+1)/2020+(x+1)/2021
Giúp mk nhanh vs nhé các bn
1 tìm x biết
x-4/2021+x-3/2020=x-2/2019+x-1/2018
giúp mk vs lm ơn
Cho x = 2020, tính giá trị:
P(x) = x^2021-2021x^2020+2021x^2019-2021x^2018+...+2021x-2020
Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Cho hàm số f(x)= x +1/4 Tính tổng f(0)+f(1/2021)+f(2/2021)+f(3/2021)+...+f(2019/2021)+f(2020/2021)+f(1)
A= x^5 - 2019 x^4 + 2019 x^3 - 2019 x^2 +2019 x -2020 tại x = 2018
a, \(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0\)
b, \(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
