Question

\(\left\{{}\begin{matrix}x^2-xy+y^2=7\left(x-y\right)\\x^2+xy+y^2=19\left(x-y\right)^2\end{matrix}\right.\)

Answer 1

Lời giải:

Nếu \(x=0\Rightarrow y=0\)

Nếu \(x\neq 0\). Đặt \(y=tx\).

\(\text{PT2}\Leftrightarrow x^2+tx^2+x^2t^2=19(x-tx)^2\)

\(\Leftrightarrow 1+t+t^2=19(1-t)^2\)

\(\Leftrightarrow 18t^2-39t+18=0\) \(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3}{2}\\t=\dfrac{2}{3}\end{matrix}\right.\)

TH1: \(t=\frac{3}{2}\Rightarrow y=\frac{3x}{2}\)

Thay vào PT (1)

\(x^2-\frac{3}{2}x^2+\frac{9}{4}x^2=7(x-\frac{3}{2}x)\)

\(\Leftrightarrow \frac{7}{4}x^2=\frac{-7}{2}x\Leftrightarrow x=-2\) (do \(x\neq 0\))

\(\Rightarrow y=-3\)

TH2: \(t=\frac{2}{3}\Rightarrow y=\frac{2}{3}x\)

Xét tương tự như TH1 ta thu được \((x,y)=(3;2)\)

Vậy \((x,y)\in \left\{(0;0);(3;2);(-2;-3)\right\}\)