Sửa đề 

\(\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left|2x-y-z\right|=0\)

Vì \(\hept{\begin{cases}\left(x-1\right)^{2018}\ge0\forall x\\\left(y+3\right)^{2020}\ge0\forall y\\\left|2x-y-z\right|\ge0\forall x,y,z\end{cases}\Rightarrow\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left|2x-y-z\right|\ge0\forall x,y,z}\)

Dấu " = " xảy ra khi :

( x - 1 )²⁰¹⁸ = 0 

=> x = 1 

( y + 3 )²⁰²⁰  = 0 

=> y = - 3 

Thay x = 1 ; y = -3 và | 2x - y - z | ta đc

| 2.1 + 3 - z | = 0 

=> | 5 - z | = 0

=> z = 5 

Vậy x = 1 ; y = -3 ; z = 5 

\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

Vì \(\left(x+y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+1\right)^2\ge0\)

\(\Rightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(\left(x+y\right)^{2018}+\left(x-2\right)^{2019}+\left(y+1\right)^{2020}=\left(1-1\right)^{2018}+\left(1-2\right)^{2019}+\left(-1+1\right)^{2020}=-1\)

Lời giải:

Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$

$\Rightarrow x=2018a; y=2019a; z=2020a$

$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$

Mặt khác:

$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$

Từ $(1); (2)$ ta có đpcm.

Ta có : (7x - 5y)²⁰¹⁸ + (3x - 2z)²⁰²⁰ + (xy + yz + xz - 4500)²⁰¹⁸ = 0

Ta có : \(\hept{\begin{cases}\left(7x-5y\right)^{2018}\ge0\\\left(3x-2z\right)^{2020}\ge0\\\left(xy+yz+xz-4500\right)^{2018}\ge0\end{cases}}\)

 \(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+xz-4500\right)^{2018}\ge0\)

Dấu bằng xảy ra <=> 

\(\begin{cases}7x=5y\\3x=2z\\xy+yz+xz=4500\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{7}\\\frac{x}{2}=\frac{z}{3}\\xy+yz+xz=4500\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}\\\frac{x}{10}=\frac{z}{15}\\xy+yz+xz=4500\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\x+y+z=4500\end{cases}}\)

Đặt \(\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=k\Rightarrow\hept{\begin{cases}x=10k\\y=14k\\z=15k\end{cases}}\)

=> xy + yz + xz = 4500

<=> 10k.14k + 14k.15k + 10k.15k = 4500

=> 140.k² + 210.k² + 150.k² = 4500

=> k².(140 + 210 + 150) = 4500

=> k² . 500 = 4500

=> k² = 9

=> k = \(\pm3\)

Nếu k = 3

=> \(\hept{\begin{cases}x=30\\y=42\\z=45\end{cases}}\)

Nếu k = - 3

=> \(\hept{\begin{cases}x=-30\\y=-42\\z=-45\end{cases}}\)

\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)