РЕШИТЕ ,ПОЖАЛУЙСТА, СИСТЕМЫ УРАВНЕНИЙ 1)x2-xy=-1 y-x=1 2)x=y-2 x2+y2=4. Created by AlexandraBragina. algebra-ru

РЕШИТЕ ,ПОЖАЛУЙСТА, СИСТЕМЫ УРАВНЕНИЙ 1)x2-xy=-1 y-x=1 2)x=y-2 x2+y2=4...

AlexandraBragina @AlexandraBragina

July 2022 1 3 Report

РЕШИТЕ ,ПОЖАЛУЙСТА, СИСТЕМЫ УРАВНЕНИЙ
1)x2-xy=-1
y-x=1
2)x=y-2
x2+y2=4

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Ответ:

1) {(1; 2)}

2) {(-2; 0), (0; 2)}

Объяснение:

$\tt \displaystyle 1) \; \left \{ {{x^{2} -x \cdot y=-1} \atop {y-x=1}} \right. \Leftrightarrow \left \{ {{x^{2} -x \cdot (1+x)=-1} \atop {y=1+x}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{x^{2} -x-x^{2}=-1} \atop {y=1+x}} \right. \Leftrightarrow \left \{ {{-x=-1} \atop {y=1+x}} \right. \Leftrightarrow \left \{ {{x=1} \atop {y=1+1=2}} \right. \Leftrightarrow \{ (1; 2) \}$

$\tt \displaystyle 2) \; \left \{ {{ x=y-2} \atop {x^{2}+y^{2}=4}} \right. \Leftrightarrow \left \{ {{ x=y-2} \atop {(y-2)^{2}+y^{2}=4}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{ x=y-2} \atop {y^{2}-4 \cdot y+4+y^{2}=4}} \right. \Leftrightarrow \left \{ {{ x=y-2} \atop {2\cdot y^{2}-4 \cdot y=0}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{ x=y-2} \atop {y^{2}-2 \cdot y=0}} \right. \Leftrightarrow \left \{ {{ x=y-2} \atop {y \cdot (y}-2)=0}} \right. \Leftrightarrow$

$\tt \displaystyle \Leftrightarrow \left \{ {{ x_{1}=0-2=-2, \; x_{2}=2-2=0} \atop {y_{1}=0, \; \; y_{2}=2}} \right. \Leftrightarrow \{(-2; 0), (0; 2) \}$

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