Question

$left { {{ sqrt{ x^{2} -2x+1}= 2y-1 } atop {|x|+y^{2}=1 }} right.$

Answer 1

$left { {{sqrt{x^2-2x+1}=2y-1} atop {|x|+y^2=1}} right. ; left { {{sqrt{(x-1)^2}=2y-1} atop {|x|+y^2=1}} right. ; left { {{|x-1|=2y-1} atop {|x|+y^2=1}} right. \\ left { {{y=frac{1}{2}(|x-1|+1)} atop {|x|+frac{1}{4}(|x-1|^2+2|x-1|+1)=1}} right. ,; ; |a|^2=a^2\\x^2-2x+1+2|x-1|+1+4|x|=1\\x^2-2x+2|x-1|+4|x|=0; ; ; ; ----(0)----(1)----\\1); x leq 0,; x^2-2x-2x+2-4x=0,\\x^2-8x+2=0,; D/4=16-2=12,\\x_1=4-2sqrt3>0,; x_2=4+2sqrt3>0\\net; reshenij$

$2); 0<x<=1,$

$x^2-2x-2x+2+4x=0,x^2+2=0,netreshenij,t.k. ; x^2+2>0$
$3); x>1,; ; x^2-2x+2x-2+4x=0,; x^2+4x-2=0,\\D/4=4+2=6,; x_1=-2-sqrt6<1,\\x_2=-2+sqrt6approx -2+2,45=0,45<1\\net; reshenij$

Вообще. система  имеет смысл при 

  $left { {{x^2-2x+1 geq 0,} atop {2y-1 geq 0,}} right. ; left { {{(x-1)^2 geq 0} atop {y geq frac{1}{2}}} right. ; left { {{xin Z} atop {y geq frac{1}{2}}} right.$