\(x^3+y^3=3xy-1\)

\(\Leftrightarrow x^3+y^3-3xy+1=0\)

\(\Leftrightarrow x^3+y^3+3x^2y+3xy^2-3xy-3x^2y-3xy^2+1=0\)

\(\Leftrightarrow\left(x+y\right)^3+1-3xy\left(x+y+1\right)=0\)

\(\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1\right)-3xy\left(x+y+1\right)=0\)

\(\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x+y+1=0\\x^2+y^2-xy-x-y+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+y=-1\\x^2+y^2-xy-x-y+1=0\end{cases}}\)

Mà x, y dương nên \(x+y=-1\)là vô lí

Vậy \(x^2+y^2-xy-x-y+1=0\)

Đến đây đợi tớ nghĩ tiếp :v

X³ + Y³ =3XY - 1

=> X³ + Y³ + 3X²Y + 3XY² - 3X²Y - 3XY² - 3XY + 1 = 0

=> \(\subset X+Y\supset^3\)+ 1 - 3XY\(\subset X+Y+1\supset\)= 0

=> \(\subset X+Y+1\supset.\)\(\subset\subset X+Y\supset^2-X-Y+1\supset\)-3XY\(\subset X+Y+1\supset=0\)

=>\(\subset X+Y+1\supset.\)\(\subset X^2+Y^2+2XY-X-Y+1-3XY\supset\)=0

=> \(\subset X+Y+1\supset.\subset X^2+Y^2-XY-X-Y+1\)=0

Vì X,Y > 0 =>X+Y+1 > 0

 \(\Rightarrow X^2+Y^2-XY-X-Y+1=0\)

\(\Rightarrow2X^2+2Y^2-2XY-2X-2Y+2=0\)

\(\Rightarrow X^2-2XY+Y^2+X^2-2X+1+Y^2-2Y+1=0\)

\(\Rightarrow\subset X-Y\supset^2+\subset X-1\supset^2+\subset Y-1\supset^2=0\)

Vì \(\subset X-Y\supset^2\ge;\subset X-1\supset^2\ge0;\subset Y-1\supset^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\subset X-Y\supset^2=0\\\subset X-1\supset^2=0\\\subset Y-1\supset^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}X-Y=0\\X-1=0\\Y-1=0\end{cases}}\)\(\Rightarrow X=Y=1\) \(\Rightarrow A=1+1=2\)

\(x^3+y^3=3xy-1\Leftrightarrow x^3+y^3+1-3xy=0\\ \)

\(\Leftrightarrow\)\(\left(x^3+3x^2y+3xy^2+y^3\right)+1-3x^2y-3xy^2-3xy=0\)

\(\Leftrightarrow\)\(\left(x+y\right)^3+1^3-3xy\left(x+y+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+y+1\right)\left(\left(x+y\right)^2-x-y+1-3xy\right)=0\)

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

\(\Leftrightarrow\)\(x^2+y^2+1^2-x-y-xy=0\)( Vì x+y+1 (> bằng 1) )

\(\Leftrightarrow\)\(2x^2+2y^2+2.1^2-2x-2y-2xy=0\)

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

\(\Leftrightarrow\)\(\hept{\begin{cases}x-y=0\\x-1=0\\y-1=0\end{cases}}\)(Do bình phương của 1 số thì luôn ( > bằng ) 0

\(\Leftrightarrow\)\(\hept{\begin{cases}x=1\\y=1\end{cases}}\)

Thay vào A, ta có:

\(A=x^{2018}+y^{2019}=1^{2018}+1^{2019}=1+1=2\)

( CTV mà ngu vl )

Nhận xét : ( x + y - 3 )^2018 >=0 và 2018.(2x-4)^2020 >= 0

=> (x+y-3)^2018 + 2018.(2x-4)^2020 >=0 

Dấu = xảy ra khi : x + y - 3 = 0 và 2x - 4 = 0 => x = 2 và y = 1

Thay vào bt S :

S = ( 2 - 1)^2019 + (2-1)^2019

= 1^2019 + 1^2019 = 2

em cảm ơn

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

Áp dụng BĐT Cô si ta có:

\(x^3+8y^3+1\ge3\sqrt[3]{x^3\cdot8y^3\cdot1}=6xy\)

\(\Rightarrow x^3+8y^3+1-6xy\ge0\)

Dấu "=" xảy ra tại \(x=2y=1\Rightarrow x=1;y=\frac{1}{2}\)

Khi đó:

\(A=x^{2018}+\left(y-\frac{1}{2}\right)^{2019}=1^{2018}+0^{2019}=1\)