Question

Найдите корни уравнения 3\|x-3| = x-4 |x-3| - знаменатель 3 - числитель

Answer 1

$\frac{3}{|x-3|}=x-4, \\ x \neq 3, \\ \left [ {{ \left \{ {{x<3, \atop {\frac{3}{3-x}=x-4,}} \right. } \atop {\left \{ {{x>3, \atop {\frac{3}{x-3}=x-4,}} \right.}} \right. \left [ {{ \left \{ {{x<3, \atop {\frac{3}{3-x}-(x-4)=0,}} \right. } \atop {\left \{ {{x>3, \atop {\frac{3}{x-3}-(x-4)=0,}} \right.}} \right. \left [ {{ \left \{ {{x<3, \atop {\frac{3-(3-x)(x-4)}{3-x}=0,}} \right. } \atop {\left \{ {{x>3, \atop {\frac{3-(x-3)(x-4)}{x-3}=0,}} \right.}} \right.$
$\left [ {{ \left \{ {{x<3, \atop {3-(3x-12-x^2+4x)=0,}} \right. } \atop {\left \{ {{x>3, \atop {3-(x^2-4x-3x+12)=0,}} \right.}} \right. \left [ {{ \left \{ {{x<3, \atop {3-7x+12+x^2=0,}} \right. } \atop {\left \{ {{x>3, \atop {3-x^2+7x-12=0,}} \right.}} \right. \left [ {{ \left \{ {{x<3, \atop {x^2-7x+15=0,}} \right. } \atop {\left \{ {{x>3, \atop {x^2-7x+9=0,}} \right.}} \right. \\ x^2-7x+15=0, \\ D=-11<0, \\ x\in\varnothing; \\ x^2-7x+9=0, \\ D=13, \\ x_1=\frac{7-\sqrt{13}}{2}\approx1,7<3,$
$x_2=\frac{7+\sqrt{13}}{2}\approx5,3, \\ x=\frac{7+\sqrt{13}}{2}.$