Question

корень(3x+1)-корень(x-1)=2

Answer 1

ОДЗ: $left { {{3x+1 geq 0} atop {x-1 geq 0}} right. left { {{x geq - frac{1}{3} } atop {x geq 1}} right. Rightarrow x geq 1$

$sqrt{3x+1} - sqrt{x-1} = 2 \ \ sqrt{3x+1} = 2+ sqrt{x-1} \ \ ( sqrt{3x+1})^2 = (2+ sqrt{x-1})^2 \ \ 3x+1 = 4 + 2sqrt{x-1} + x-1 \ \ x-1 = 2sqrt{x-1} \ \ (x-1)^2 = (2sqrt{x-1} )^2 \ \ x^2 - 2x+1 = 4x-4 \ \ x^2-6x+5 = 0$
Корни уравнения 
$x_{1} = 5 ; x_{2} = 1$

Ответ:  $x_{1} = 5 ; x_{2} = 1$