\(\Leftrightarrow\left(1\right)-\left(2\right)=x^2-y^2=0\\ \Leftrightarrow\left(x-y\right)\left(x+y\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
Với \(x=y\)
\(\left(1\right)\Leftrightarrow x^2-x^2=28\left(vn\right)\)
Với \(x=-y\)
\(\left(1\right)\Leftrightarrow x^2+x^2=28\Leftrightarrow x^2=14\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\Leftrightarrow y=-\sqrt{14}\\x=-\sqrt{14}\Leftrightarrow y=\sqrt{14}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(\sqrt{14};-\sqrt{14}\right);\left(-\sqrt{14};\sqrt{14}\right)\right\}\)
Lời giải:
Lấy $(1)-(2)$ ta thu được:
$x^2-y^2=0$
$\Leftrightarrow (x-y)(x+y)=0$
$\Rightarrow x-y=0$ hoặc $x+y=0$
Nếu $x-y=0\Leftrightarrow x=y$. Thay vô (1):
$x^2-x^2=28\Leftrightarrow 0=28$ (vô lý)
Nếu $x+y=0\Leftrightarrow x=-y$. Thay vô (2):
$(-y)^2-(-y)(y)=28$
$\Leftrightarrow 2y^2=28\Leftrightarrow y^2=14\Rightarrow y=\pm \sqrt{14}$
$\Rightarrow x=\mp \sqrt{14}$ (tương ứng)
Vậy $(x,y)=(\sqrt{14}, -\sqrt{14}), (-\sqrt{14}, \sqrt{14})$
