x² + 2y² + z² - 2xy - 2y - 4z + 5 = 0

<=> ( x² - 2xy + y² ) + ( y² - 2y + 1 ) + ( z² - 4z + 4 ) = 0

<=> ( x - y )² + ( y - 1 )² + ( z - 2 )² = 0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )² + ( y - 1 )² + ( z - 2 )²\(\ge\)0\(\forall\)x ; y ; z

Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )

Thay ( 1 ) vào A , ta được :

\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)

Vậy A = 2

Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)

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

Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)

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

Lời giải:
Nếu $x\geq 1$ thì $2^x$ chẵn

$\Rightarrow 2^x+5999$ lẻ

$\Rightarrow 4y$ lẻ (vô lý)

Do đó $x<1$. Mà $x$ tự nhiên nên $x=0$

$4y=2^x+5999=2^0+5999=6000$

$\Rightarrow y=1500$ 

Vậy $x=0; y=1500$

$(x-1)^{2019}+(y-1501)^{2020}=(0-1)^{2019}+(1500-1501)^{2020}$

$=(-1)+1=0$

ĐKXĐ: \(\left\{{}\begin{matrix}2020-y^2\ge0\\2020-z^2\ge0\\2020-x^2\ge0\end{matrix}\right.\)

Ta có:

\(x\sqrt{2020-y^2}+y\sqrt{2020-z^2}+z\sqrt{2020-x^2}=3030\)

\(\Leftrightarrow2x\sqrt{2020-y^2}+2y\sqrt{2020-z^2}+2z\sqrt{2020-x^2}=6060\)

\(\Leftrightarrow2020-y^2-2x\sqrt{2020-y^2}+x^2+2020-z^2-2y\sqrt{2020-z^2}+y^2+2020-x^2-2z\sqrt{2020-x^2}+z^2=0\)

   \(\Leftrightarrow\left(\sqrt{2020-y^2}-x\right)^2+\left(\sqrt{2020-z^2}-y\right)^2+\left(\sqrt{2020-x^2}-z\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{2020-y^2}-x\right)^2=\left(\sqrt{2020-z^2}-y\right)^2=\left(\sqrt{2020-x^2}-z\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2020-y^2}=x\\\sqrt{2020-z^2}=y\\\sqrt{2020-x^2}=z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2020-y^2=x^2\\2020-z^2=y^2\\2020-x^2=z^2\end{matrix}\right.\)(vì \(x,y,z>0\))

\(\Leftrightarrow\left\{{}\begin{matrix}2020=x^2+y^2\\2020=y^2+z^2\\2020=z^2+x^2\end{matrix}\right.\)

\(\Rightarrow2\left(x^2+y^2+z^2\right)=3.2020\)

\(\Rightarrow x^2+y^2+z^2=3.1010=3030\)

\(\Rightarrow A=x^2+y^2+z^2=3030\)

Vậy \(A=3030\)

Ta có: (x - 2)⁴ \(\ge\)0 \(\forall\)x

          (2y - 1)²⁰²⁰ \(\ge\) 0 \(\forall\)y

=> (x - 2)⁴ + (2y - 1)²⁰²⁰ \(\ge\)0 \(\forall\)x,y

Mà ĐK : (x - 2)⁴ + (2y - 1)²⁰²⁰ \(\le\)0

=> (x - 2)⁴ + (2y - 1)²⁰²⁰ = 0

=> \(\hept{\begin{cases}\left(x-2\right)^4=0\\\left(2y-1\right)^{2020}=0\end{cases}}\)

=>  \(\hept{\begin{cases}x-2=0\\2y-1=0\end{cases}}\)

=> \(\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Với x = 2, y = 1/2 thay vào biểu thức P, ta có:

   P = \(21.2^2.\frac{1}{2}+4.2.\left(\frac{1}{2}\right)^2\) = \(42+2=44\)

           Vậy giá trị của P = 44

x+y=1=>y=1-x

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

\(=\left(x^2+2x+1\right)+\left(x+\frac{1}{x}\right)+2018\)\(=\left(x+1\right)^2+\left(x+\frac{1}{x}\right)+2018\)

Ta có: \(\left(x+1\right)^2\ge0\forall x>0\)

Áp dụng BĐT Cô-si cho 2 số dương \(x\)và \(\frac{1}{x}\):

\(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)

\(\Rightarrow Q\ge2+2018=2020\)

Dấu '=' xảy ra \(\Leftrightarrow\hept{\begin{cases}x+1=0\\x=\frac{1}{x}\end{cases}\Leftrightarrow x=-1}\)\(\Rightarrow y=1-\left(-1\right)=2\)

Vậy \(minQ=2020\Leftrightarrow x=-1;y=2\)

    Ta có : (x-2014)^2010 >=0 và (y-2010)^2014 >= 0 nên:

                  (x-2014)^2010 + ( y-2010)^2014 >=0

           Dấu bằng xảy ra khi:

                    (x-2014) ^2010=0 và (y-2010)^2014 =0

           Suy ra :  (x-2014)=0 và (y-2010)=0

            =>        x=2014 và y=2010     => x+y = 2014+2010=4024