Ta có: \(x^3+y^3+z^3=3xyz\)

   \(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)

   \(\Leftrightarrow\left(x+y+z\right)^3-3.\left(x+y\right).z.\left(x+y+z\right)-3xy\left(x+y\right)-3xyz=0\)

   \(\Leftrightarrow\left(x+y+z\right).\left[\left(x+y+z\right)^2-3.\left(x+y\right).z\right]-3xy\left(x+y+z\right)=0\)

   \(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2zx-3xz-3yz-3xy\right)=0\)

   \(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2-xz-yz-xy\right)=0\)

+ \(x+y+z=0\)\(\Rightarrow\)\(C=\frac{x^{2019}+y^{2019}+z^{2019}}{0}\)( Loại )

+ \(x^2+y^2+z^2-xz-yz-xy=0\)

\(\Rightarrow2x^2+2y^2+2z^2-2xz-2yz-2xy=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow\)\(x=y=z\)

\(\Rightarrow\)\(C=\frac{x^{2019}+x^{2019}+x^{2019}}{\left(x+x+x\right)^{2019}}=\frac{3.x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)

Vậy.......

Từ x³ + y³ + z³ = 3xyz

=> ( x + y + z )( x² + y² + z² - xy - yz - xz ) = 0 ( phân tích như bạn kia )

Vì x + y + z ≠ 0

=> x² + y² + z² - xy - yz - xz = 0

<=> 2x² + 2y² + 2z² - 2xy - 2yz - 2xz = 0

<=> ( x - y )² + ( y - z )² + ( x - z )² = 0

VT ≥ 0 ∀ x,y,z. Đẳng thức xảy ra <=> x=y=z

Khi đó \(C=\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}=\frac{3x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}\cdot x^{2019}}=\frac{1}{3^{2018}}\)

Ta có : x³ + y³ = z(3xy - z²)

=> x³ + y³ = 3xyz - z³

=> x³ + y³ + z³ - 3xyz = 0

=> (x + y)(x² - xy + y²) + z³ - 3xyz = 0

=> (x + y)³ - 3xy(x + y) + z³ - 3xyz = 0

=> [(x + y)³ + z³] - 3xy(x + y) - 3xyz  = 0

=> (x + y + z)[(x + y)² - (x + y)z + z²] - 3xy(x + y + z) = 0

=> (x + y +z)(x² + y ² + 2xy - xz - yz + z²) - 3xy(x + y + z) = 0

=> (x + y + z)(x² + y² + z² - xy - yz - zx) = 0

=> x² + y² + z² - xy - yz - zx = 0 (Vì x + y + z = 3)

=> 2(x² + y² + z² - xy - yz - zx) = 0

=> 2x² + 2y² + 2z² - 2xy - 2yz - 2zx = 0

=> (x² - 2xy + y²) + (y² - 2yz + z²) + (x² - 2zx + z²) = 0

=> (x - y)² + (y - z)² + (x - z)² = 0

=> \(\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}}\Rightarrow x=y=z\)

mà x + y + z = 3

=> x = y = z = 1

Khi đó A = 673(x²⁰¹⁹ + y²⁰¹⁹ + z²⁰¹⁹) + 1 

= 673(1²⁰¹⁹ + 1²⁰¹⁹ + 1²⁰¹⁹) + 1

= 673.3 + 1 = 2020

Vậy A = 2020

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)

\(\Rightarrow A=0\)

Sai đề không

Sai đề thật, cái biểu thức trên 100% lớn hơn hoặc = 9, lấy đâu ra =1

Ta có : \(A=\frac{2019}{x+xy+1}+\frac{2019}{y+yz+1}+\frac{2019}{z+zx+1}=2019\left(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+xyz+z}+\frac{xz}{xyz+xyz^2+xz}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+z+1}+\frac{xz}{1+z+xz}+\frac{1}{z+zx+1}\right)\)(vì xyz = 1)

\(=2019\left(\frac{z+xz+1}{xz+z+1}\right)=2019\)

Vậy A = 2019