Question

Найдите сумму корней уравнения v3-x + v6+x=3 (v - корень). буду : )

Answer 1

$\sqrt{3-x}+ \sqrt{6+x} =3\\ ========================\\ \left \{ {{3-x \geq 0} \atop {6+x \geq 0}} \right. \\ \left \{ {{-x \geq -3} \atop {x \geq -6}} \right. \\ \left \{ {{x \leq 3} \atop {x \geq -6}} \right. \Rightarrow x \in [-6;3] \\ ========================\\ (\sqrt{3-x}+ \sqrt{6+x})^2 =3^2\\ 3-x+2 \sqrt{(3-x)(6+x)}+6+x=9\\ 2 \sqrt{(3-x)(6+x)}=0\\ \sqrt{(3-x)(6+x)}=0\\ 3-x=0 \rightarrow x=3\\ 6+x=0 \rightarrow x=-6$

ответ: x1 = 3; x2 = -6