Решите уравнениеА) x^2+3x+2/2-x =oБ) x+4=5/xВ) x^2-13x^2+36=0

Решите задачу:

$\frac{x^2+3x+2}{2-x}=0,~~~x\neq 2\\x^2+3x+2=0\\x^2+2x+x+2=0\\x(x+2)+x+2)=0\\(x+2)(x+1)=0\\x+2=0\\x+1=0\\x_1=-2\\x_2=-1$

$x+4=\frac{5}{x},~~~~~x\neq 0\\(x+4)*x=5\\x^2+4x-5=0\\x^2+5x-x-5=0\\x(x+5)-(x-5)=0\\(x+5)(x-1)=0\\x+5=0\\x-1=0\\x_1=-5\\x_2=1$

$x^2-13x^2+36=0\\-12x^2+36=0\\x^2-3=0\\x^2=3\\x_1=-\sqrt{3}\\x_2=\sqrt{3}$

Решите уравнениеА) x^2+3x+2/2-x =oБ) x+4=5/xВ) x^2-13x^2+36=0​