Решите систему уравнений x+y=1 x^2+y^2=17

[latex] \left \{ {{x+y=1} \atop {x^2+y^2=17}} \right. ; \left \{ {{y=1-x} \atop {x^2+(1-x)^2=17}} \right. ; \left \{ {{y=1-x} \atop {x^2+1-2x+x^2=17}} \right. ; \left \{ {{y=1-x} \atop {2x^2-2x-16=0}} \right. ;[/latex] [latex] \left \{ {{y=1-x} \atop {x^2-x-8=0}} \right. ;[/latex] [latex]D=(-1)^2-4*1*(-8)=1+32=33[/latex] [latex]x_{1,2}= \frac{1\pm \sqrt{33} }{2} [/latex] [latex]y_{1,2}=1- \frac{1\pm \sqrt{33} }{2} = \frac{2-(1\pm \sqrt{33} )}{2} = \frac{1\mp \sqrt{33}}{2} [/latex] имеем две точки: [latex](\frac{1\pm \sqrt{33}}{2} ;\frac{1\mp \sqrt{33}}{2} )[/latex] Ответ: [latex](\frac{1+ \sqrt{33}}{2} ;\frac{1- \sqrt{33}}{2} );(\frac{1- \sqrt{33}}{2} ;\frac{1+ \sqrt{33}}{2} )[/latex]

если х+у = 1,  то (х+у)² = 1 = х² + у² + 2ху 2ху = -16 ху = -8 х = -8/у -8/у + у = 1 у² - у - 8 = 0 D=1+32 = 33 y₁ = (1-√33)/2   --->   x₁ = -8*2/(1-√33) = -16(1+√33)/(-32) = (1+√33)/2 y₂ = (1+√33)/2   --->   x₂ = -8*2/(1+√33) = -16(1-√33)/(-32) = (1-√33)/2 ПРОВЕРКА: (1+√33)/2 + (1-√33)/2 = (1+√33+1-√33)/2 = 2.2 = 1 (1+√33)²/4 + (1-√33)²/4 = (1+2√33+33+1-2√33+33)/4 = 68/4 = 17