Question

[Nguyễn Thị Ngọc Hân]

Giải hệ pt

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

b) \(\left\{{}\begin{matrix}2x^2-3xy+y^2=3\\x^2+2xy-2y^2=6\end{matrix}\right.\)

Answer 1

[Nguyễn Việt Lâm]

a/ \(\Leftrightarrow\left\{{}\begin{matrix}4x^2-16xy+4y^2=4\\y^2-3xy=4\end{matrix}\right.\)

\(\Rightarrow4x^2-13xy+3y^2=0\)

\(\Leftrightarrow\left(x-3y\right)\left(4x-y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3y\\y=4x\end{matrix}\right.\)

Thay vào pt sau: \(\left[{}\begin{matrix}y^2-3y.y=4\left(vn\right)\\\left(4x\right)^2-3x.4x=4\end{matrix}\right.\)

\(\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1;y=4\\x=-1;y=-4\end{matrix}\right.\)

b/

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)

\(\Rightarrow3x^2-8xy+4y^2=0\)

\(\Leftrightarrow\left(x-2y\right)\left(3x-2y\right)=0\Rightarrow\left[{}\begin{matrix}x=2y\\x=\frac{2}{3}y\end{matrix}\right.\)

Thay vào pt đầu: \(\left[{}\begin{matrix}2\left(2y\right)^2-3.2y.y+y^2=3\\2\left(\frac{2}{3}y\right)^2-3.\frac{2}{3}y.y+y^2=3\end{matrix}\right.\) bạn tự giải nốt