Question

Решите уравнения:
(3-x)(x+4)+x^2=0
(1-x)(2-x)=(x+3)(x-4)

Answer 1

$(3-x)(x+4)+x^2=0\\3x+12-x^2-4x+x^2\\-x+12=0\\-x=-12\\x=12\\\\(1-x)(2-x)=(x+3)(x-4)\\2-x-2x+x^2=x^2-4x+3x-12\\2-x-2x=-4x+3x-12\\2-3x=-x-12\\-3x+x=-12-2\\-2x=-14\\x=(-14):(-2)\\x=7$