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x = 2020 => 2021 = x + 1
x2020 - 2021x2019 + 2021x2018 - 2021x2017 + ... + 2021x2 - 2021x + 1
= x2020 - ( x + 1 )x2019 + ( x + 1 )x2018 - ( x + 1 )x2017 + ... + ( x + 1 )x2 - ( x + 1 )x + 1
= x2020 - x2020 - x2019 + x2019 + x2018 - x2018 - x2017 + ... + x3 + x2 - x2 - x + 1
= -x + 1 = -2020 + 1 = -2019
Vậy giá trị của biểu thức = -2019
f(2020) = 20206 - 2021 × 20205 + 2021 × 20204 - 2021×20203 + 2021×20202 - 2021 × 2020 + 2021 = 1
Chúc bn học tốt !!!!!!!
Ta có: \(\left|x+\frac{1}{2021}\right|\ge0\) ; \(\left|x+\frac{2}{2021}\right|\ge0\) ; ... ; \(\left|x+\frac{2020}{2021}\right|\ge0\) \(\left(\forall x\right)\)
\(\Rightarrow\left|x+\frac{1}{2021}\right|+\left|x+\frac{2}{2021}\right|+...+\left|x+\frac{2020}{2021}\right|\ge0\left(\forall x\right)\)
\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)
Từ đó ta được: \(x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)
\(\Leftrightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)
\(\Leftrightarrow x=\frac{\left(2020+1\right)\left[\left(2020-1\right)\div1+1\right]}{2021}\)
\(\Leftrightarrow x=\frac{2021\cdot2020}{2021}=2020\)
Vậy x = 2020
\(\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|=2021x\)
Ta có:\(\left|\frac{x+1}{2021}\right|\ge0;\left|\frac{x+2}{2021}\right|\ge0;....;\left|\frac{x+2020}{2021}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|\ge0\forall x\)
\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\frac{x+1}{2021}+\frac{x+2}{2021}+...+\frac{x+2020}{2021}=2021x\)
\(\Rightarrow x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)
\(\Rightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)
\(\Rightarrow x=2020\)
a) Có x = 2020 => x + 1 = 2021. Thay 2021 = x + 1 vào A
\(A=x^6-\left(x+1\right)^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)
\(A=1\)
b) Có x = -19 => x - 1 = -20 => - ( x - 1 ) = 20. Thay 20 = - ( x - 1) vào B
\(B=x^{10}-\left(x-1\right)x^9-\left(x-1\right)x^8-\left(x-1\right)x^7-...-\left(x-1\right)x^2-\left(x-1\right)x-x+1\)
\(B=x^{10}-x^{10}+x^9-x^9+...+x^2-x^2+x-x+1\)
\(B=1\)
Chúc bạn học tốt!!!
Đặt \(A\left(x\right)=x^2-2021x+2020=0\)
\(\Leftrightarrow x^2-2020x-x+2020=0\)
\(\Leftrightarrow x\left(x-1\right)-2020\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-2020\right)\left(x-1\right)=0\Leftrightarrow x=\orbr{\begin{cases}x=2020\\x=1\end{cases}}\)
Vậy nghiệm của phương trình là x = 1 ; x = 2020
Ta có: \(\hept{\begin{cases}\left(2021x-1\right)^{2020}\ge0\\\left(3y+4\right)^{2022}\ge0\end{cases}}\left(\forall x,y\right)\)
\(\Rightarrow\left(2021x-1\right)^{2020}+\left(3y+4\right)^{2022}\ge0\left(\forall x,y\right)\)
Mà theo đề bài ta có: \(\left(2021x-1\right)^{2020}+\left(3y+4\right)^{2022}\le0\)
Nên từ đó suy ra: \(\hept{\begin{cases}\left(2021x-1\right)^{2020}=0\\\left(3y+4\right)^{2022}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2021x-1=0\\3y+4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2021}\\y=-\frac{4}{3}\end{cases}}\)
Khi đó \(M=2021\cdot\frac{1}{2021}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(=-\frac{4}{3}-\frac{16}{9}=-\frac{28}{9}\)
Ta có : \(x=2022\Rightarrow x-1=2021\)
hay \(B=x^{10}-\left(x-1\right)x^9-\left(x-1\right)x^8-...-\left(x-1\right)x^2-\left(x-1\right)x+5\)
\(=x^{10}-x^{10}+x^9-x^9+x^8-...-x^3+x^2-x^2+x+5\)
\(=x+5\Rightarrow B=2022+5=2027\)
Vậy với x = 2022 thì B = 2027
x=2020 nên x+1=2021
\(P\left(x\right)=x^{2021}-x^{2020}\left(x+1\right)+x^{2019}\left(x+1\right)-....+x\left(x+1\right)-2020\)
\(=x^{2021}-x^{2021}-x^{2020}+x^{2020}-...+x^2+x-2020\)
=x-2020=0