Question

Решите систему уравнений:
x^2+y^2=13,
x^4-y^4=65.​

Answer 1

Ответ:

(3; 2), (-3; 2), (3; -2), (-3; -2) .

Объяснение:

$\left \{\begin{array}{l} x^{2} +y^{2} = 13, \\ x^{4} -y^{4} = 65; \end{array} \right.\Leftrightarrow \left \{\begin{array}{l} x^{2} +y^{2} = 13, \\ (x^{2} -y^{2})(x^{2} +y^{2} ) = 65; \end{array} \right.\Leftrightarrow \left \{\begin{array}{l} x^{2} +y^{2} = 13, \\ 13(x^{2} -y^{2} ) = 65; \end{array} \right.\Leftrightarrow$

$\Leftrightarrow \left \{\begin{array}{l} x^{2} +y^{2} = 13, \\ (x^{2} -y^{2} ) = 65:13; \end{array} \right.\Leftrightarrow\Leftrightarrow \left \{\begin{array}{l} x^{2} +y^{2} = 13, \\ x^{2} -y^{2} = 5; \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} x^{2} +y^{2} = 13, \\ 2x^{2} = 18; \end{array} \right.\Leftrightarrow$

$\Leftrightarrow \left \{\begin{array}{l} x^{2} +y^{2} = 13, \\ x^{2} = 9 \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} 9+y^{2} = 13, \\ x^{2} = 9 \end{array} \right.\Leftrightarrow\left \{\begin{array}{l}y^{2} = 4, \\ x^{2} = 9 \end{array} \right.\Leftrightarrow$

$\left \{\begin{array}{l} \left[\begin{array}{l} y = 2, \\ y = -2 \end{array} \right. \\\\ \left [\begin{array}{l} x= 3, \\ x = -3 \end{array} \right.\end{array} \right.\Leftrightarrow\left [\begin{array}{l} \left \{\begin{array}{l} x= 3, \\ y= 2; \end{array} \right. \\ \left \{\begin{array}{l} x = -3, \\ y = 2; \end{array} \right. \\ \left \{\begin{array}{l} x = 3, \\ y = -2; \end{array} \right.\\ \left \{\begin{array}{l} x = -3, \\ y = -2. \end{array} \right. \end{array} \right.$

Тогда система имеет четыре решения (3; 2), (-3; 2), (3; -2), (-3; -2)