[tex] \left \{ {{ \sqrt{ x^{2} -2x+1}= 2y-1 } \atop {|x|+y^{2}=1 }} \right. [/tex]. Created by Kolonizator80. algebra-ru

tex]...

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[tex] \left \{ {{ \sqrt{ x^{2} -2x+1}= 2y-1 }
\atop {|x|+y^{2}=1 }} \right. [/tex]

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NNNLLL54

Verified answer

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

$2)\; 0$

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

Вообще. система имеет смысл при

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

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