tìm x biết :|x+1/2021|+|x+2/2021|+...+|x+2020/2021|=2021x
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CG
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XO
7 tháng 3 2021
Ta có x = 2020
=> x + 1 = 2021
A = x2021 - 2021x2020 + .... + 2021x - 2021
= x2021 - (x + 1)x2020 + .... + (x + 1)x - (x + 1)
= x2021 - x2021 - x2020 + .... + x2 + x - x + 1
= 1
Vậy A = 1
7 tháng 3 2021
Ta có : \(x=2020\Rightarrow x+1=2021\)
\(A=x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}-\left(x+1\right)x^{2018}+...-\left(x+1\right)x^2+\left(x+1\right)x-2021\)
= x2021 - x2021 - x2020 + x2020 + x2019 - x2019 - x2018 + ... - x3 - x2 + x2 + x - 2021 = x - 2021
mà x = 2020 hay 2020 - 2021 = -1
Vậy với x = 2020 thì A = -1
7 tháng 9 2021
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
14 tháng 4 2022
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
HP
8 tháng 10 2021
2021 - x + 2021(x - 2020x) = 0
<=> 2021 - x + 2021 - 4082420 = 0
<=> -x - 4082420 = 0
<=> x = -4082420
HP
1
BL
0
BL
1
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:\(\left|\frac{x+1}{2021}\right|\ge0;\left|\frac{x+2}{2021}\right|\ge0;....;\left|\frac{x+2020}{2021}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|\ge0\forall x\)
\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\frac{x+1}{2021}+\frac{x+2}{2021}+...+\frac{x+2020}{2021}=2021x\)
\(\Rightarrow x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)
\(\Rightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)
\(\Rightarrow x=2020\)