\(ĐK:y\ne0\)

\(x+y=\dfrac{x}{y}\Leftrightarrow xy+y^2=x\)

Mà \(xy=x+y\Leftrightarrow x+y+y^2=x\)

\(\Leftrightarrow y\left(y+1\right)=0\Leftrightarrow y=-1\left(y\ne0\right)\\ \Leftrightarrow x-1=\dfrac{x}{-1}=-x\\ \Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(P=2022\cdot\dfrac{1}{2}+2021\left(-1\right)=1011-2021=-1010\)

x^2+y^2-2x-4y+6=1-(x-y+1)^2

=>x^2-2x+1+y^2-4y+4=-(x-y+1)^2

=>(x-1)^2+(y-2)^2=-(x-y+1)^2

=>(x-1)^2+(y-2)^2+(x-y+1)^2=0

=>x=1;y=2

A=2022+2023*2

=2022+4046

=6068

( x - 1 )²⁰¹⁸ + (y - 2 )²⁰²⁰+(z-3)²⁰²²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)

\(A=\dfrac{1}{9}\left(-x\right)^{2021}y^2z^3=\dfrac{1}{3}\left(-1\right)^{2021}.2^2.3^3=\dfrac{1}{3}.\left(-1\right).4.27=-36\)

Biến đổi: 4 x 2 − 4 xy + y 2 = 0 ⇔ ( 2 x − y ) 2 = 0 ⇔ 2 x = y  

Thay y = 2x vào P ta được P = -3

\(P=\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}+2020=\dfrac{x^5+y^5}{\left(xy\right)^2}+2020=\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)-\left(xy\right)^2\left(x+y\right)}{\left(-2\right)^2}\)

\(=\dfrac{\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\left[\left(x+y\right)^2-2xy\right]-\left(-2\right)^2.5}{4}\)

\(=\dfrac{\left(-8+6.5\right)\left(25+4\right)-20}{4}=...\)

\(A=\dfrac{x^2+y^2}{xy}-4xy=\dfrac{\left(x+y\right)^2-2xy}{xy}-4xy\)

\(=\dfrac{4x^2y^2-2xy}{xy}-4xy=4xy-2-4xy=-2\)