Question

Помогите пожалуйста(75баллов)
(Задание прикреплено) ​

Answer 1

$\displaystyle 3.\\\\1). \ 7m-n+49m^{2}-n^{2} = 7m-n+(7m-n)(7m+n)=\\\\{} \ \ \ \ (7m-n)(7m+n+1);\\\\2). \ 4x^{2}-4xy+y^{2}-16=(2x-y)^{2}-16=(2x-y-4)(2x-y+4);\\\\3). \ xy^{4}-2y^{4}-xy+2y=y^{4}(x-2)-y(x-2)=y(y^{3}-1)(x-2)=\\\\{} \ \ \ \ y(y-1)(y^{2}+y+1)(x-2);\\\\4). \ 9-x^{2}-2xy-y^{2}=9-(x+y)^{2}=(3-x-y)(3+x+y)=\\\\{} \ \ \ \ =-(x+y-3)(x+y+3);$

$\displaystyle 4.\\\\1). \ 5x^{3}-5x=0\\\\{} \ \ \ \ 5x(x-1)(x+1)=0\\\\{} \ \ \ \ x_{1}=0 \ \ \ \ \ \ \ x_{2}=1 \ \ \ \ \ \ \ x_{3}=-1\\\\2). \ 64x^{3}-16x^{2}+x=0\\\\{} \ \ \ \ x((8x)^{2}-16x+1)=0\\\\{} \ \ \ \ x(8x-1)^{2}=0\\\\{} \ \ \ \ x_{1}=0 \ \ \ \ \ \ \ x_{2}=\frac{1}{8}\\\\3). \ x^{3}-3x^{2}-4x+12=0\\\\{} \ \ \ \ x^{2}(x-3)-4(x-3)=0\\\\{} \ \ \ \ (x-2)(x+2)(x-3)=0\\\\{} \ \ \ \ x_{1}=2 \ \ \ \ \ \ \ \ x_{2}=-2 \ \ \ \ \ \ \ \ x_{3}=3$