Question

Giải pt

\(\left\{{}\begin{matrix}x^2+2xy-3y^2=9\\2x^2+2xy+y^2=2\end{matrix}\right.\)

Answer 1

\(HPT\Leftrightarrow\left\{{}\begin{matrix}2x^2+4xy-6y^2=18\left(1\right)\\18x^2+18xy+9y^2=18\left(2\right)\end{matrix}\right.\)

Lấy \(PT\left(2\right)\) trừ đi \(PT\left(1\right)\) ta có : \(16x^2+14xy+15y^2=0\)

\(\Leftrightarrow\left(4x\right)^2+2.4x.\dfrac{7}{4}y+\dfrac{49}{16}y^2+\dfrac{191}{16}y^2=0\Leftrightarrow\left(4x+\dfrac{7}{4}y\right)^2+\dfrac{191}{16}y^2=0\)

\(\Rightarrow\left(x;y\right)=\left(0;0\right)\left(loại\right)\)

Vậy hệ PT vô nghiệm