1) $\left\{\begin{matrix} x^{3}+12x=y^{3}+12y & \\ x^{4}+y^{4}=16 & \end{matrix}\right.$
2
1) $\left\{\begin{matrix} x^{3}+12x=y^{3}+12y & \\ x^{4}+y^{4}=16 & \end{matrix}\right.$
2
Hệ phương trình <=> $\left\{\begin{matrix}(x^{3}-y^{3})+12(x-y)=0 \\ x^{4}+y^{4}=16 \end{matrix}\right. <=> \left\{\begin{matrix}(x-y)(x^{2}+xy+y^{2}+12)=0 \\ x^{4}+y^{4} \end{matrix}\right.$
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