Question

Розв'яжіть систему рівнянь.

Answer 1

Объяснение:

$\displaystyle\\\left \{ {{\sqrt{x^2-5}+\sqrt{y^2+5}=7 } \atop {x^2+y^2=29}} \right.\ \ \ \ \ \left \{ {{\sqrt{x^2-5}+\sqrt{y^2+5}=7 } \atop {(x^2-5)+(y^2+5)=29}} \right. .$

Пусть: $\sqrt{x^2-5}=t\ \ \ \ \ \ \sqrt{y^2+5}=v\ \ \ \ \ \ \Rightarrow$

$\displaystyle\\\left \{ {{t+v=7} \atop {t^2+v^2=29}} \right. \ \ \ \ \ \ \left \{ {{v=7-t} \atop {t^2+(7-t)^2=29}} \right. \ \ \ \ \ \ \left \{ {{v=7-t} \atop {t^2+49-14t+t^2=29}} \right. \\\\\\\left \{ {{v=7-t} \atop {2t^2-14t+20=0\ |:2}} \right. \ \ \ \ \ \ \left \{ {{v=7-t} \atop {t^2-7t+10=0}} \right. \ \ \ \ \ \ \left \{ {{v=7-t} \atop {t^2-5t-2t+10=0}} \right.$

$\displaystyle\\\left \{ {{v=7-t} \atop {t*(t-5)-2*(t-5)=0}} \right. \ \ \ \ \ \ \left \{ {{v=7-t} \atop {(t-5)*(t-2)=0}} \right. \ \ \ \ \ \ \left \{ {{v_1=2\ \ \ \ v_2=5} \atop {t_1=5\ \ \ \ t_2=2}} \right. .\\\\\\1)\\\\\left \{ {{\sqrt{y^2+5}=2 } \atop {\sqrt{x^2-5}=5 }} \right.\ \ \ \ \ \ \left \{ {{(\sqrt{y^2+5})^2=2^2} \atop {(\sqrt{x^2-5})^2=5^2}} \right. \ \ \ \ \ \ \left \{ {{y^2+5=4} \atop {x^2-5=25}} \right.\ \ \ \ \ \ \left \{ {{y^2=-1\notin} \atop {x^2=30}} \right. \notin .$

$2)\\ \displaystyle\\ \left \{ {{\sqrt{y^2+5}=5 } \atop {\sqrt{x^2-5}=2 }} \right. \ \ \ \ \ \ \left \{ {(\sqrt{y^2+5})^2=5^2} } \atop {(\sqrt{x^2-5})^2 =2^2}} \right. \ \ \ \ \ \ \left \{ {{y^2+5=25} \atop {x^2-5=4}} \right.\ \ \ \ \ \ \left \{ {{y^2=20} \atop {x^2=9}} \right. \\\\\\\left \{ {{y_1=-2\sqrt{5} \ \ \ \ y_2=2\sqrt{5} } \atop {x_1=-3\ \ \ \ x_2=3}} \right. .$