Question

(x2 + y2 = 81; x2 - y = 9
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Answer 1

Ответ:

{(0; -9), (-$\sqrt{17}$; 8), ($\sqrt{17}$; 8)}

Объяснение:

$\displaystyle \tt \left \{ {{x^2+y^2=81} \atop {x^2-y=9}} \right. \Leftrightarrow \left \{ {{x^2+y^2=81} \atop {x^2=y+9}} \right. \Leftrightarrow \left \{ {{y+9+y^2=81} \atop {x^2=y+9}} \right. \Leftrightarrow \left \{ {{y^2-81+y+9=0} \atop {x^2=y+9}} \right. \Leftrightarrow$

$\displaystyle \tt \Leftrightarrow \left \{ {{(y+9) \cdot (y-9)+y+9=0} \atop {x^2=y+9}} \right. \Leftrightarrow \left \{ {{(y+9) \cdot (y-8)=0} \atop {x^2=y+9}} \right. \Leftrightarrow \left \{ {{y_1=-9, y_2=8} \atop {x^2=y+9}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{y_1=-9, y_2=8} \atop {x^2=y_1+9=-9+9=0, x^2=y_2+9=8+9=17}} \right. \Leftrightarrow$

$\displaystyle \tt \Leftrightarrow \left \{ {{y_1=-9, y_2=8} \atop {x_{1}=0, x_{2,3}=\mp \sqrt{17} }} \right. \Leftrightarrow (x; y) \in \{ (0; -9), (-\sqrt{17}; 8 ), (\sqrt{17}; 8) \}.$