Question

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

Answer 1

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