Под корнем 1-3х = х+3

[latex] \sqrt{1-3x} = x+3 [/latex] [latex] (\sqrt{1 - 3x})^2 = (x+3)^2 [/latex] [latex]1 - 3x = x^2 + 6x + 9 [/latex] [latex] x^2 + 9x + 8 = 0 [/latex] [latex]D = b^2 - 4ac = 81 - 4*1*8 = 49[/latex] [latex]x_{1}= \frac{-9 + 7}{2} = -1[/latex] [latex] x_{2} = \frac{-9-7}{2} = -8[/latex] Проверка:  x= -1  [latex] \sqrt{1 + 3} = -1+3[/latex] верно x = -8 [latex] \sqrt{1 + 24} = -8+3[/latex] неверно  Ответ: х = -1

ОДЗ [latex] \left \{ {{1-3x \geq 0} \atop {x+3 \geq 0}} \right. \ \ \left \{ {{3x \leq 1} \atop {x\geq -3}} \right.\ \ \ \left \{ {{x \leq \frac{1}{3}} \atop {x\geq -3}} \right.[/latex] x∈[[latex]-3; \frac{1}{3}[/latex]] [latex]\sqrt{1-3x} = x+3 \\ (\sqrt{1-3x})^2 = (x+3)^2 \\ 1-3x=x^2+6x+9 \\ x^2+9x+8=0 \\ D=81-32=49 \\ x_1=\frac{-9+7}{2}=-1 \ \ \ \ x_2=\frac{-9-7}{2}=-8[/latex] x₂ не удовлетворяет ОДЗ Ответ x=-1