x=2
y=3
z=4

\(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}=\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2z-4x=0\\4y-3z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\\ \Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x-2y+3z}{2-6+12}=\dfrac{8}{8}=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\\z=4\end{matrix}\right.\)

a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)

b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)

\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\) ; \(4y=3z\Rightarrow\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow\frac{x}{2}=\frac{2y}{6}=\frac{3z}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\frac{x}{2}=\frac{2y}{6}=\frac{3z}{12}=\frac{x+2y+3z}{2+6+12}=\frac{-20}{20}=-1\)

\(\Rightarrow x=-2;y=-3;z=-4\)

a: \(A=11x^4y^3z^2+20x^2yz-4xy^2z+10x^2yz-3x^4y^3z^2-2008xyz^2-8x^4y^3z^2\)

\(=30x^2yz-4xy^2z-2008xyz^2\)

b: \(A=30x^2yz-4xy^2z-2xyz\left(15x-2y\right)\)

\(=30x^2yz-4xy^2z-30x^2yz+4xy^2z=0\)

Bậc của A là 9.