tính nhanh
2022 x5 _ 5 x 1
2021 4 4 2021
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2020 + 1 2020 - 2021 - 1 2021 x - 1 2 = 1 3 - 2 15 - 1 5 ( 2020 + 1 2020 - 2021 - 1 2021 ) ( x - 1 2 ) = 5 15 - 2 15 - 3 15 ( 2020 + 1 2020 - 2021 - 1 2021 ) ( x - 1 2 ) = 0
Mà
( 2020 + 1 2020 - 2021 - 1 2021 ) = ( 1 2020 - 1 2021 ) + 2020 - 2021
1 2020 . 2021 - 1 < 0
Do 1 2020 . 2021 < 1
\(M=\left(x^5-2021x^4\right)-\left(x^4-2021x^3\right)+\left(x^3-2021X^2\right)-\left(x^2-2021x\right)+\left(x-2021\right)-900=-900\)
Ta có: x=2021
nên x+1=2022
Ta có: \(M=x^5-2022x^4+2022x^3-2022x^2+2022x-2921\)
\(=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-2921\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2921\)
\(=x-2921=-900\)
= 3/4 + 1/3 = 13/12
5/4 x X = 5/8
X = 5/8 : 5/4
X = 1/2
vậy X = ...
\(x=2021\Leftrightarrow x+1=2022\\ \Leftrightarrow P=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x\\ P=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x\\ P=0\)
\(P=x^5-2022x^4+2022x^3-2022x^2+2022x-2021=x^4\left(x-2021\right)-x^3\left(x-2021\right)+x^2\left(x-2021\right)-x\left(x-2021\right)+\left(x-2021\right)\)
\(=\left(x-2021\right)\left(x^4-x^3+x^2-x+1\right)\)
\(=\left(2021-2021\right)\left(x^4-x^3+x^2-x+1\right)=0\)