\(\left(x-3\right)^{2020}-\left(x-3\right)^{2021}=0\\ \left(x-3\right)^{2020}-\left(x-3\right)^{2020}.\left(x-3\right)=0\\ \left(x-3\right)^{2020}\left(1-x+3\right)=0\\ \left(x-3\right)^{2020}\left(4-x\right)=0\\ \left[{}\begin{matrix}\left(x-3\right)^{2020}=0\\4-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Tìm x thuộc Z : x^3 + x = 2021^2020
x-4/2022+x-3/2021+x-2/2020+x-1/2019=-4
Nếu 1/3 + 1/6 +1/10 + ...... + 1/x.(x+1) : 2 = 2019/2021
A.x = 2019/2020 B. x = 2019 C. x = 2020 D. x = 2021
*Thực hiện
1/ (\(\dfrac{2021}{2020}\)+\(\dfrac{2020}{2021}\)) x (\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\))
2/ (\(\dfrac{7}{19}\)-\(\dfrac{5}{12}\)):\(\dfrac{-5}{8}\)-(\(\dfrac{7}{19}\)-\(\dfrac{29}{12}\)):\(\dfrac{5}{8}\)
3/ \(\dfrac{-5}{6}\)x\(\dfrac{7}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{14}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{3}{24}\)
\(\dfrac{x+1}{2021}\) + \(\dfrac{x+2}{2020}\) + \(\dfrac{x+1}{2019}\) = -3
Tìm x bt:
\(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}+\dfrac{x+3}{2019}\)
Tìm x
a)x.(x+2021)=0
b)(x-2020).(x+2021)=0
c)(x-2021).(x2+1)=0
d)(x+1)+(x+3)(x+5)+.....+(x+99)=0
2019 x 2020 - 1/ 2019 x 2020 và 2020 x 2021 - 1/ 2020 x 2021
so sánh phân số
So sánh:
A=2021^2020+2/2021^2020-1 và B=2021^2020/2021^2020-3
tìm số nguyên tố x sao cho
x3+x=20212020
