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Ответ:

[tex]\displaystyle (x_1;y_1)=(\frac{-1+\sqrt{7} }{2} ;\frac{1+\sqrt{7} }{2});\\ (x_2;y_2)=(\frac{-1-\sqrt{7} }{2} ;\frac{1-\sqrt{7} }{2});[/tex]

Объяснение:

[tex]\displaystyle \left \{ {{x^2+y^2=4} \atop {y-x=1}} \right. < = > \left \{ {{x^2+(1+x)^2=4} \atop {y=1+x}} \right. < = > \left \{ {{x^2+1+2x+x^2-4=0} \atop {y=1+x}} \right. < = > \\ < = > \left \{ {{2x^2+2x-3=0} \atop {y=1+x}} \right.[/tex]
Решим отдельно первое уравнение системы
2x²+2x-3 = 0;
D = 2²-4*2*(-3) = 4+24 = 28 = (√4*7)² = (2√7)²
[tex]\displaystyle x_{12} =\frac{-2 \pm 2\sqrt{7} }{2*2}[/tex]
[tex]\displaystyle x_{1} =\frac{-2 + 2\sqrt{7} }{2*2}=\frac{-1+\sqrt{7} }{2} ;\\x_{2} =\frac{-2 - 2\sqrt{7} }{2*2}=\frac{-1-\sqrt{7} }{2}[/tex]
Найдём у:
[tex]\displaystyle y_{1} =1+\frac{-1+\sqrt{7} }{2} =\frac{2-1+\sqrt{7} }{2}=\frac{1+\sqrt{7} }{2} \\y_{2} =1+\frac{-1-\sqrt{7} }{2}=\frac{2-1-\sqrt{7} }{2}=\frac{1-\sqrt{7} }{2}[/tex]