Đáp án: $(x,y)\in\{(\dfrac15, -1), (5, 1)\}$

Giải thích các bước giải:

Ta có:

$\begin{cases}13x(y^2+1)=5(x^2+y^2)\\ 13y(x^2-1)=12(x^2+y^2)\end{cases}$

$\to \begin{cases}\dfrac{y^2+1}{x^2+y^2}=\dfrac{5}{13x}\\ \dfrac{x^2-1}{x^2+y^2}=\dfrac{12}{13y}\end{cases}$

$\to \begin{cases}\dfrac{y^2+1}{x^2+y^2}+\dfrac{x^2-1}{x^2+y^2}=\dfrac{5}{13x}+\dfrac{12}{13y}\\ \dfrac{x^2-1}{x^2+y^2}=\dfrac{12}{13y}\end{cases}$

$\to \begin{cases}1=\dfrac{5}{13x}+\dfrac{12}{13y}\\ \dfrac{x^2-1}{x^2+y^2}=\dfrac{12}{13y}\end{cases}$

$\to \begin{cases}\dfrac{12}{13y}=1-\dfrac{5}{13x}\\ \dfrac{x^2-1}{x^2+y^2}=\dfrac{12}{13y}\end{cases}$

$\to \begin{cases}\dfrac{12}{13y}=\dfrac{13x-5}{13x}\\ \dfrac{x^2-1}{x^2+y^2}=\dfrac{12}{13y}\end{cases}$

$\to \begin{cases}y=\dfrac{12x}{13x-5}\\ \dfrac{x^2-1}{x^2+(\dfrac{12x}{13x-5})^2}=\dfrac{12}{13\cdot \dfrac{12x}{13x-5}}\end{cases}$

$\to \begin{cases}y=\dfrac{12x}{13x-5}\\ x\in\{5, \dfrac15\}\end{cases}$

$\to (x,y)\in\{(\dfrac15, -1), (5, 1)\}$