Để đơn giản, đặt \(\left(a;-2b;3c\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\x^2+y^2+z^2=18\end{matrix}\right.\)
Ta cần tính \(P=x^4+y^4+z^4\)
\(xy+yz+zx=\frac{\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)}{2}=-9\)
\(\Rightarrow2\left(x^2y^2+y^2z^2+z^2x^2\right)=\left(xy+yz+zx\right)^2-2xyz\left(x+y+z\right)=81\)
\(x^4+y^4+z^4=\frac{\left(x^2+y^2+z^2\right)^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)}{2}=\frac{18^2-81}{2}=\frac{243}{2}\)
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