Question

x^2-(V2+1)x+V2=0
ПОМОГИТЕ СРОЧНО 25 БАЛЛОВ

Answer 1

$x^{2} -(\sqrt{2}+1)x+\sqrt{2}=0\\x^{2}-\sqrt{2}x-x+\sqrt{2}=0\\x^{2} +(-\sqrt{2}-1)x+\sqrt{2}=0\\D=(\sqrt{2}+1)^2-4\sqrt{2}=(-\sqrt{2})^2-2\sqrt{2}+1=(-\sqrt{2}+1)^2\\x1=\frac{\sqrt{2}+1+\sqrt{2}-1}{2}=\sqrt{2}\\x2=\frac{\sqrt{2}+1-\sqrt{2}+1}{2}=1$

Ответ: $1;$ $\sqrt{2}$

$x^{2} -(\sqrt{2}+1)x+\sqrt{2}=0\\x1+x2=\sqrt{2}+1\\x1x2=\sqrt{2}\\x1=\sqrt{2}\\x2=1$

Ответ: $1;$ $\sqrt{2}$