Ta có: \((x + y + z)^2 = 0 \)
\(\Leftrightarrow\) \( x^2 + y^2 + z^2 + 2.(xy + yz + xz) = 0 \)
\(\Leftrightarrow\) \(1 + 2(xy + yz + xz) = 0 \)
\(\Leftrightarrow\) \(xy + yz + xz =\) \(-\dfrac{1}{2}\)
Lại có:
\(x^2y^2 + y^2z^2 + x^2z^2 \)
\(= (xy + yz + xz)^2 - 2xyz(x + y + z) \)
\(=\dfrac{1}{4}-0\)
\(=\dfrac{1}{4}\)
\(\Rightarrow\) \( x^4+y^4+z^4 = (x^2 + y^2 + z^2)^2 - 2(x^2y^2 + y^2z^2 + x^2z^2) \)
\(=\) \(1 - 2.\dfrac{1}{4} \)
\(=\dfrac{1}{2}\)
Vậy \(x^4+y^4+z^4=\dfrac{1}{2}\)
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