решить кто решит (28x³y+27x-25ab³)(23x²-12ac²+2x)​

27x

3

−y

3

=(3x−y)(9y

2

+3xy+y

2

)

−3x

2

−12x−12=−3(x

2

+4x+4)=−3(x+2)

2

−625=−1∗(25)

2

=−1∗(5)

4

25a

3

−ab

2

=a(5a−b)(5a+b)

3ab−15a+12b−60=3a(b−5)+12(b−5)=(b−5)(3a+12)

2,

x(x-1)(x+1)-(x-2)(x^{2} +2x+4)=x^{3}-x)-(x-2)(x+2)(x+2)=x^{3}-x-(x+2)(x^{2} -4)=x^{3}-x-(x^{3}+2x^{2} -4x-8)=x^{3}-x-x^{3}-2x^{2} +4x+8=-2x^{2} +3x+8x(x−1)(x+1)−(x−2)(x

2

+2x+4)=x

3

−x)−(x−2)(x+2)(x+2)=x

3

−x−(x+2)(x

2

−4)=x

3

−x−(x

3

+2x

2

−4x−8)=x

3

−x−x

3

−2x

2

+4x+8=−2x

2

+3x+8

3.

\begin{gathered}y^{3} + 18y^{2} + 81y=y(y^{2} +18y+81)=y(y+9)^{2} \\ x - 2- xy + 2y=(x-2)-y(x-2)=(x-2)(1-y)\\4x^{2} -4xy+y^{2} -16=(2x-y)^{2}-16=(2x-y-4)(2x-y+4)\end{gathered}

y

3

+18y

2

+81y=y(y

2

+18y+81)=y(y+9)

2

x−2−xy+2y=(x−2)−y(x−2)=(x−2)(1−y)

4x

2

−4xy+y

2

−16=(2x−y)

2

−16=(2x−y−4)(2x−y+4)

4

\begin{gathered}5x^{3} - 5x = 0\\5x(x^{2} -1)=0\\5x(x-1)(x+1)=0\\\left[\begin{array}{ccc}5x=0\\x-1=0\\x+1=0\end{array}left[\begin{array}{ccc}x_{1} =0\\x_{2}=1 \\x_{3} =-1\end{array}\end{gathered} \begin{gathered}x³ - 3x² - 4x + 12 = 0\\x(x^{2} -4)-3(x^{2} -4)=0\\(x^{2} -4)(x-3)=0\\(x-2)(x+2)(x-2)=0\\\left[\begin{array}{ccc}x-2=0\\x+2=0\\\end{array}left \{ {{x_{1} =2} \atop {x_{2} =-2}} \right.\end{gathered} \begin{gathered}64x^{3} - 16x^{2} + x = 0\\x((8x)^{2} )-2*8x+1)=0\\x(8x-1)^{2} =0\\\left \{ {{x_{1} =0} \atop {x_{2} =\frac{1}{8} }} \right.\end{gathered}

64x

3

−16x

2

+x=0

x((8x)

2

)−2∗8x+1)=0

x(8x−1)

2

=0

{

x

2

=

8

1

x

1

=0