ta có 

\(x^2+y^2+z^2\)\(=200\)

\(2xy-yz-zx=M\)

\(\Leftrightarrow M+200=x^2+y^2+z^2+2xy-yz-zx\)

\(\Leftrightarrow M+200=\left(x+y\right)^2-z\left(x+y\right)+z^2\)

\(\Leftrightarrow\left(x+y-\frac{z}{2}\right)^2+\frac{3}{4}z^2\ge0\)

\(\Leftrightarrow M\ge-200\)

\(\left(x^3+3x^2y+3xy^2+y^3-z^3\right):\left(x+y-z\right)\\ =\left[\left(x+y\right)^3-z^3\right]:\left(x+y-z\right)\\ =\left(x+y-z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]:\left(x+y-z\right)\\ =x^2+2xy+y^2+xz+yz+z^2\)

Vậy chọn A 

\(M+200=x^2+y^2+z^2+2xy-yz-xz\ge0\)

\(\Leftrightarrow x^2+x\left(2y-z\right)+y^2+z^2-yz\ge0\)

Can cm \(\left(2y-z\right)^2-4\left(y^2+z^2-yz\right)\le0\)

\(\Leftrightarrow3z^2\ge0\). TU dok ta co \(M+200\ge0\rightarrow M\ge-200\)

\("="\Leftrightarrow\left(x;y;z\right)=\left(10;-10;0\right)=\left(-10;10;0\right)\)

a+b+c=3 nha (quên bổ sung)

Bài 1:

$a^2+b^2+c^2=ab+bc+ac$
$\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0$

$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$

Vì $(a-b)^2, (b-c)^2, (c-a)^2\geq 0$ với mọi $a,b,c$

Do đó để tổng của chúng bằng $0$ thì $a-b=b-c=c-a=0$

$\Leftrightarrow a=b=c$

Mà $a+b+c=3$ nên $a=b=c=1$

$\Rightarrow Q=(1+1)^2+(1+2)^3+(1+3)^3=95$

Lấy 3 lần pt dưới cộng pt trên ta được :
\(4x^2+4y^2+z^2+2yz-4xz-4xy=0\)

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

\(\Leftrightarrow\hept{\begin{cases}y=0\\2x-y-z=0\end{cases}\Rightarrow\hept{\begin{cases}y=0\\z=2x\end{cases}}}\)

\(\Rightarrow x^2+4x^2-2x^2=3\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=1;z=2\\x=-1;z=-2\end{cases}}\)