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mình thấy tội nghiệp hai bạn đó mà =(

\(2x^2+2y^2+2xy-4x+4y+8=0\)

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

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

\(\rightarrow x=-y=2\)

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\(x^4+4x^3-5x^2-38x-30=0\)

\(\Leftrightarrow x^4-2x^3-3x^2+6x^3-12x^2-18x+10x^2-20x-30=0\)

\(\Leftrightarrow x^2\left(x^2-2x-3\right)+6x\left(x^2-2x-3\right)+10\left(x^2-2x-3\right)=0\)

\(\Leftrightarrow\left(x^2-2x-3\right)\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow\left(x^2-3x+x-3\right)\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow\left[x\left(x-3\right)+\left(x-3\right)\right]\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x^2+6x+10\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\\x^2+6x+10=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x^2+6x+9+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\\left(x+3\right)^2+1>0\left(loai\right)\end{matrix}\right.\)

Từ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow\left\{{}\begin{matrix}1+\dfrac{x}{y}+\dfrac{x}{z}=0\left(1\right)\\1+\dfrac{y}{x}+\dfrac{y}{z}=0\left(2\right)\\1+\dfrac{z}{x}+\dfrac{z}{y}=0\left(3\right)\end{matrix}\right.\)

Và \(\dfrac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)

\(\Rightarrow\left(xy+yz+xz\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=0\)

\(\Rightarrow\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{y}{z}=0\)

\(\Rightarrow A+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{y}{z}=0\)

Cộng theo vế của \(\left(1\right);\left(2\right);\left(3\right)\)suy ra:

\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}=-3\)

\(\Rightarrow A-3=0\Rightarrow A=3\)