Question

помогите пж с примером

Answer 1

$\displaystyle \frac{x^2+3x}{x-4}=\frac{x^2-x}{4-x},x\neq 4\\ \\(x^2+3x)(4-x)=(x^2-x)(x-4)\\ (x^2+3x)(4-x)-(x^2-x)(x-4)=0\\4x^2-x^3+12x-3x^2-(x^3-4x^2-x^2+4x)=0\\4x^2-x^3+12x-3x^2-(x^3-5x^2+4x)=0\\4x^2-x^3+12x-3x^2-x^3+5x^2-4x=0\\6x^2-2x^3+8x=0\\2x(3x-x^2+4)=0\\2x(-x^2+3x+4)=0\\2x(-x^2+4x-x+4)=0\\2x(-x(x-4)-(x-4))=0\\2x(-(x-4)(x+1))=0\\-2x(x-4)(x+1)=0\\x(x-4)(x+1)=0\\x=0,x-4=0,x+1=0\\x=0,x=4,x=-1\\x_1=-1,x_2=0$