Question

с2 и с3 пожалуйста, с решением. заранее спасибо

Answer 1

Ответ:

2) 2

3) (x1,y1)=(0;5)

(x2,y2)=(5;-20)

Answer 2

С2

$\sqrt{(6-\sqrt{20})^2}+\sqrt{(4-2\sqrt{5})^2}=|6-\sqrt{20}|+|4-2\sqrt{5}|=\\\\=(6-\sqrt{4\cdot5})-(4-2\sqrt{5})=6-2\sqrt{5}-4+2\sqrt{5}=2\\\\ OTBET:2$

С3

$-\left \{ {{x^2+y=5} \atop {5x+y=5}} \right. \\ \\ \left \{ {{(x^2+y)-5x+y=5-5} \atop {5x+y=5}} \right. \\ \\ \left \{ {{x^2-5x=0} \atop {y=5-5x}} \right. \\ \\ \left \{ {{x(x-5)=0} \atop {y=5-5x}} \right. \\ \\ \left[\begin{array}{ccc}\left \{ {{x=0} \atop {y=5-5\cdot0}} \right. \\ \left \{ {{x=5} \atop {y=5-5\cdot5}}\end{array}$

$\left[\begin{array}{ccc}\left \{ {{x=0} \atop {y=5}} \right. \\ \left \{ {{x=5} \atop {y=-20}}\end{array}$

$OTBET:(0;5),(5;-20)$