Question

под корнем 1-3х = х+3

Answer 1

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

Answer 2

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