Question

Решите срочно пожалуйста!!!!!!
1 вариант

Answer 1

Ответ:

$1)\ \ \dfrac{x^2}{2-x}=\dfrac{3x}{2-x}\ \ ,\ \ \ ODZ:\ x\ne 2\ ,\\\\x^2=3x\ \ ,\ \ x^2-3x=0\ \ ,\ \ x\, (x-3)=0\\\\x_1=0\ \ ,\ \ x_2=3\\\\\\2)\ \ \dfrac{x+1}{x-3}+\dfrac{12}{x+3}=\dfrac{24}{x^2-9}\ \ ,\ \ \ \ ODZ:\ x\ne \pm 3\ ,\\\\\\\dfrac{(x+1)(x+3)+12(x-3)-24}{(x-3)(x+3)}=0\\\\\\x^2+4x+3+12x-36-24=0\\\\x^2+16x-57=0\ \ ,\ \ D/4=121\ ,\ \ x_1=-19\ \ ,\ \ x_2=3\notin ODZ\\\\Otvet:\ \ x=-19\ .$

$3)\ \ \dfrac{x+5}{x-2}+\dfrac{x-2}{x+1}=\dfrac{21}{(x-2)(x+1)}\ \ ,\ \ \ \ ODZ:\ x\ne 2\ ,\ x\ne -1\ ,\\\\\\\dfrac{(x+5)(x+1)+(x-2)^2-21}{(x-2)(x+1)}=0\\\\\\x^2+6x+5+x^2-4x+4-21=0\\\\x^2+2x-12=0\ \ ,\ \ D/4=13\ ,\ \ x_{1,2}=-1\pm \sqrt{13}$