Chủ đề này có 1 trả lời

[hieu31320001]

$\left\{\begin{matrix} & 3y\sqrt{x+2}+8\sqrt{x+2}=10y-3xy+12\\ & 5y^{3}\sqrt{2-x}-8=6y^{2}+xy^{3}\sqrt{2-x} \end{matrix}\right.$

[didifulls]

$\left\{\begin{matrix} & 3y\sqrt{x+2}+8\sqrt{x+2}=10y-3xy+12\\ & 5y^{3}\sqrt{2-x}-8=6y^{2}+xy^{3}\sqrt{2-x} \end{matrix}\right.$

pt $(2)\Leftrightarrow y^3\sqrt{2-x}(3+2-x)=6y^2+8$.

$\Leftrightarrow (\sqrt{2-x})^3+3(\sqrt{2-x})=(\frac{2}{y})^3+3\frac{2}{y}.$