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:

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

Sửa đề: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)

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

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

=>x=y=1 và z=2

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

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

=1

ta có x^2 + 2y^2 +z^2 -2xy -2y -4z +5 =0

=> (x^2 - 2xy +y^2) + (y^2 -2y +1) + (z^2 -4z +4) =0

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

=> x=y , y=1 , z=2 

=> A= (1-1)^2018 + (1-1)^2019 + ( 2-1)^2020 => A= 1

nghĩ thế !

Tham khảo:

CHÚC EM HỌC TỐT NHÁ 

Bài 1 :

Vì \(\sqrt{3x+2y+z}\ge0\forall x;y;z\)

\(\left|y-\frac{1}{2}\right|\ge0\forall y\)

\(\left(z-2\right)^2\ge0\forall z\)

\(\Rightarrow A\ge2018\forall x;y;z\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x+2y+z=0\\y-\frac{1}{2}=0\\z-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x+2\cdot\frac{1}{2}+2=0\\y=\frac{1}{2}\\z=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=\frac{1}{2}\\z=2\end{cases}}}\)

Vậy........

Bài 2 :

Lý luận tương tự câu 1) ta có :

\(\hept{\begin{cases}x-1=0\\y+1=0\\x+y+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\\1-1+z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\\z=0\end{cases}}}\)

Thay x; y; z vào P ta có :

\(P=1^{2018}+\left(-1\right)^{2019}+0^{2020}\)

\(P=1-1+0\)

\(P=0\)

3x^2+3y^2+4xy-2x+2y+2=0

=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0

=>x=1 và y=-1

M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1

Phân tích GT đầu , ta có : x = y = z

Rồi làm như thường

mình sửa đề nhé~

Có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x;y;z\)

\(\Rightarrow2.\left(x^2+y^2+z^2\right)-2xy-2yz-2xz\ge0\forall x;y;z\)

\(\Leftrightarrow2.\left(x^2+y^2+z^2\right)\ge2xy+2yz+2xz\forall x;y;z\)

\(\Leftrightarrow3.\left(x^2+y^2+z^2\right)\ge x^2+y^2+z^2+2xy+2yz+2xz\forall x;y;z\)

\(\Leftrightarrow3.\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\forall x;y;z\)

Mà \(3.\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)

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

Có: \(x^{2018}+y^{2018}+z^{2018}=27^{673}\)

\(\Leftrightarrow3.x^{2018}=27^{673}\)

\(\Leftrightarrow x^{2018}=3^{2018}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

đến đây bạn tự làm nốt nhé