Question

[poppy Trang]

giải hệ phương trình:

\(\left\{{}\begin{matrix}x\left(x^2+3x+4\right)=8y^3+2y-2\\x^3-2xy=\frac{1}{3}\end{matrix}\right.\)

Answer 1

[Nguyễn Việt Lâm]

\(x^3+3x^2+4x+2=8y^3+2y\)

\(\Leftrightarrow\left(x+1\right)^3-\left(2y\right)^3+\left(x+1-2y\right)=0\)

\(\Leftrightarrow\left(x+1-2y\right)\left[\left(x+1\right)^2+2y\left(x+1\right)+4y^2+1\right]=0\)

\(\Leftrightarrow x+1-2y=0\Rightarrow2y=x+1\)

\(\Rightarrow x^3-x\left(x+1\right)-\frac{1}{3}=0\)

\(\Leftrightarrow3x^3-3x^2-3x-1=0\)

\(\Rightarrow x=\frac{\sqrt[3]{16}+\sqrt[3]{4}+1}{3}\)