ĐKXĐ: ...
Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\ge0\)
\(\left\{{}\begin{matrix}a+b+c=4\\a^2+b^2+c^2=6\\a^4+b^4+c^4=18\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ab+bc+ca=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=5\\a^2b^2+b^2c^2+c^2a^2=\frac{\left(a^2+b^2+c^2\right)^2-\left(a^4+b^4+c^4\right)}{2}=9\end{matrix}\right.\)
\(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\)
\(\Rightarrow abc=2\)
Ta có hệ mới: \(\left\{{}\begin{matrix}a+b+c=4\\ab+bc+ca=5\\abc=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+c=4-a\\bc+a\left(b+c\right)=5\\bc=\frac{2}{a}\end{matrix}\right.\) \(\Rightarrow a\left(4-a\right)+\frac{2}{a}=5\)
\(\Leftrightarrow a^3-4a^2+5a-2=0\)
\(\Leftrightarrow\left(a-1\right)^2\left(a-2\right)=0\)
