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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:\(\left|\frac{x+1}{2021}\right|\ge0;\left|\frac{x+2}{2021}\right|\ge0;....;\left|\frac{x+2020}{2021}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|\ge0\forall x\)
\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\frac{x+1}{2021}+\frac{x+2}{2021}+...+\frac{x+2020}{2021}=2021x\)
\(\Rightarrow x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)
\(\Rightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)
\(\Rightarrow x=2020\)
\(x^2+y^2+z^2=1\Rightarrow x^2,y^2,z^2\le1\Rightarrow-1\le x,y,z\le1\)
Ta có:\(x^3+y^3+z^3-x^2-y^2-z^2=0\)
\(\Rightarrow x^2\left(x-1\right)+y^2\left(y-1\right)+z^2\left(z-1\right)=0\)
Vì \(x-1\le0,y-1\le0,z-1\le0\)
\(\Rightarrow x^2\left(x-1\right)\text{}\le0,y^2\left(y-1\right)\le0,z^2\left(z-1\right)\le0\)
\(\Rightarrow x^2\left(x-1\right)\text{}+y^2\left(y-1\right)+z^2\left(z-1\right)\le0\)
Dấu "=" xảy ra khi\(\left\{{}\begin{matrix}x^2\left(x-1\right)=0\\y^2\left(y-1\right)=0\\z^2\left(z-1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left(x,y,z\right)\) là bộ (0,0,1) và các hoán vị
\(\Rightarrow x^{2021}+y^{2021}+z^{2021}=1\)
{ x + y + z = 1 (1)
{ x² + y² + z² = 1 (2)
{ x³ + y³ + z³ = 1 (3)
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
⇒ 2(xy + yz + zx) = (x + y + z)² - (x² + y² + z²) = 1² - 1 = 0 ⇒ xy + yz + zx = 0
(x + y + z)³ = x³ + y³ + z³ + 3(x + y)(y + z)(z + x)
⇒ 3(x + y)(y + z)(z + x) = (x + y + z)³ - (x³ + y³ + z³) = 1³ - 1 = 0
⇒ x + y = 0 hoặc y + z = 0 hoặc z + x = 0
@ Nếu x + y = 0 ⇔ x = - y thay vào (1) ⇒ z = 1 , thay vào (2) ⇒ 2x² + 1 = 1 ⇒ x = 0; y = 0
⇒ S = 1
Tương tự cho trường hợp y + z = 0 và z + x = 0
x ≥ 2021
x ≥ 2021