Vũ Trúc Linh ơi  ^2 là sao nhỉ                                                                                        

có x+y=2021=>y=2021-x

=>x.y=x(2021-x)=2021x-\(x^2\)

=>P=2021x-\(x^2\)

=> -P=\(x^2-2021x\)\(=x^2-2.\dfrac{2021}{2}.x+\left(\dfrac{2021}{2}\right)^2-\left(\dfrac{2021}{2}\right)^2\)=\(\left(x-\dfrac{2021}{2}\right)^2-\left(\dfrac{2021}{2}\right)^2\)

lại có x,y nguyên dương=>x,y\(\ge\)1

có x+y=2021=>x,y\(\le\)2020

=>\(x\le2020\)

=>\(x-\dfrac{2021}{2}\le2020-\dfrac{2021}{2}\)

<=>\(\left(x-\dfrac{2021}{2}\right)^2\le\left(\dfrac{2019}{2}\right)^2\)

=>\(\left(x-\dfrac{2021}{2}\right)^2-\left(\dfrac{2021}{2}\right)^2\le\)\(\left(\dfrac{2019}{2}\right)^2-\left(\dfrac{2021}{2}\right)^2=-2020\)

<=>\(-P\le-2020< =>P\ge2020\)

dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2020\\x=1\end{matrix}\right.\)

vậy MIN P=2020 khi x=2020 hoặc x=1

bổ sung đoạn cuối dấu với x=2020 thì y=1

với x=1 thì y =2020

??????

Ta có : \(\frac{x+y\sqrt{2021}}{y+z\sqrt{2021}}=\frac{a}{b}\left(a,b\inℕ^∗;\left(a,b\right)=1\right)\)

<=>\(bx-ay=\left(az-by\right)\sqrt{2021}\)

<=>\(\hept{\begin{cases}nx-ay=0\\az-by=0\end{cases}}\)<=>\(\frac{x}{y}=\frac{y}{z}=\frac{a}{b}\)=> xz = y²

Lại có : x² + y² + z² = ( x + z )² - 2xz + y² = ( x + z )² - y² = ( x + z - y ) ( x + z + y )

Vì x + y + z > 1 và x² + y² + z² là số ntố => \(\hept{\begin{cases}x^2+y^2+z^2=x+y+z\\x-y+z=1\end{cases}}\)<=> x = y = z = 1 ( tm )