Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}4x^2-2y^2=2\\x^2+xy=2\end{matrix}\right.\)

Answer 1

\(\begin{cases}4x^2-2y^2=2\\ x^2+xy=2\end{cases}\)

=>\(4x^2-2y^2-x^2-xy=2-2=0\)

=>\(3x^2-xy-2y^2=0\)

=>\(3x^2-3xy+2xy-2y^2=0\)

=>3x(x-y)+2y(x-y)=0

=>(x-y)(3x+2y)=0

TH1: x-y=0

=>x=y

\(x^2+xy=2\)

=>\(x^2+x^2=2\)

=>\(2x^2=2\)

=>\(x^2=1\)

=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)

Nếu x=1 thì y=x=1

Nếu x=-1 thì y=x=-1

TH2: 3x+2y=0

=>3x=-2y

=>\(x=-\frac{2y}{3}\)

\(x^2+xy=2\)
=>\(\left(-\frac{2y}{3}\right)^2+y\cdot\frac{-2y}{3}=2\)

=>\(\frac{4y^2}{9}-\frac{2y^2}{3}=2\)

=>\(-\frac{2y^2}{9}=2\)

=>\(y^2=-9\) <0(loại)

Vậy: (x;y)∈{(1;1);(-1;-1)}