[tex] {x}^{2} + {y}^{2} = 8 \\ x + y = 2 [/tex]
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Ответ:

[tex](1+\sqrt{3}; 1-\sqrt{3}) \quad ; \quad (1-\sqrt{3}; 1+\sqrt{3}) \quad ;[/tex]

Объяснение:

[tex]$ \displaystyle \left \{ {{x^{2}+y^{2}=8} \atop {x+y=2}} \right. \Leftrightarrow \left \{ {{x^{2}+(2-x)^{2}=8} \atop {y=2-x}} \right. \Leftrightarrow \left \{ {{x^{2}+4-4x+x^{2}=8} \atop {y=2-x}} \right. \Leftrightarrow $[/tex]

[tex]$ \displaystyle \left \{ {{2x^{2}-4x-4=0} \atop {y=2-x}} \right. \Leftrightarrow \left \{ {{x^{2}-2x-2=0} \atop {y=2-x}} \right. ; $[/tex]

[tex]x^{2}-2x-2=0;[/tex]

[tex]D=b^{2}-4ac;[/tex]

[tex]D=(-2)^{2}-4 \cdot 1 \cdot (-2)=4+8=12=(2\sqrt{3})^{2};[/tex]

[tex]x_{1,2}=\dfrac{-b \pm \sqrt{D}}{2a} \Rightarrow x_{1,2}=\dfrac{-(-2) \pm \sqrt{(2\sqrt{3})^{2}}}{2 \cdot 1}=\dfrac{2 \pm 2\sqrt{3}}{2}=1 \pm \sqrt{3};[/tex]

[tex]x_{1,2}=1 \pm \sqrt{3} , \quad y=2-x \Rightarrow y_{1,2}=2-(1 \pm \sqrt{3})=1 \mp \sqrt{3};[/tex]

[tex](1+\sqrt{3}; 1-\sqrt{3}) \quad ; \quad (1-\sqrt{3}; 1+\sqrt{3}) \quad ;[/tex]