Question

[Vua Phá Lưới]

Giải hệ pt

Answer 1

[Trương Huy Hoàng]

\(\left\{{}\begin{matrix}x^3+y^3+x^2y+xy^2=32\\x^2y^2\left(x^2+y^2\right)=128\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2\right)=32\\x^2y^2\left(x^2+y^2\right)=128\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2y^2-4x-4y=0\\x^2y^2\left(x^2+y^2\right)=128\end{matrix}\right.\) \(\left(x,y\ne0\right)\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{2\left(1\pm\sqrt{\left(1+y\right)\left(1-y+y^2\right)}\right)}{y^2}\\x^2y^2\left(x^2+y^2\right)=128\end{matrix}\right.\)

Thay x ở pt 1 trên vào pt 2 r biến đối ta tìm được y

Mình tìm được 1 nghiệm (2;2), bạn tự tìm tiếp nha, nhưng mình nghĩ chắc hết r

Chúc bn học tốt!

Answer 2

[Nguyễn Việt Lâm]

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-2xy\left(x+y\right)=32\\x^2y^2\left[\left(x+y\right)^2-2xy\right]=128\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}u^3-2uv=32\\u^2v^2-2v^3=128\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u\left(u^2-2v\right)=32\\v^2\left(u^2-2v\right)=128\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u\left(u^2-2v\right)=21\\\dfrac{v^2}{u}=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-2uv=32\\u=\dfrac{v^2}{4}\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{v^2}{4}\right)^3-\dfrac{2v^3}{4}=32\)

\(\Rightarrow\dfrac{v^6}{64}-\dfrac{v^3}{2}-32=0\) \(\Rightarrow\left[{}\begin{matrix}v^3=64\\v^3=-32\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}v=4\Rightarrow u=4\\v=-\sqrt[3]{32}\Rightarrow u=8\sqrt[3]{2}\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=4\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=8\sqrt[3]{2}\\xy=-\sqrt[3]{32}\end{matrix}\right.\)