2x(x+3)+2(x+3)=0
=> (x+3)(2x+2)=0
=> 2(x+3)(x+1)=0

TH1: x+3=0 => x = -3

TH2: x+1=0 => x=-1

\(2x.\left(x+3\right)+2.\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right).\left(2x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0-3\\2x=0-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-1\end{cases}}\)

ta có : đa thức

\(-x^3+6x^2-x+a\text{ chia hết cho x-1}\) khi nó cũng có nghiệm x=1

vậy ta có :

\(-1^3+6.1^2-1+a=0\Leftrightarrow a=-4\)

áp dụng hằng đẳng thức: \(a^3+b^3=\left(a+b\right).\left(a^2-ab+b^2\right)\)

\(\left(8y^3+8\right):\left(4y^2-4y+4\right)\)

\(=[\left(2y\right)^3+2^3]:\left(4y^2-4y+4\right)\)

\(=\left(2y+2\right).[\left(2y\right)^2-2y.2+2^2]:\left(4y^2-4y+4\right)\)

\(=\left(2y+2\right).\left(4y^2-4y+4\right):\left(4y^2-4y+4\right)\)

\(=2y+2\)

a) (x-4).x-(x-3)^2=0

=>x^2-4x-(x^2-2.x.3+9)=0

=>x^2-4x-x^2+6x-9=0

=>(x^2-x^2)-4x+6x-9=0

=>2x-9=0

=>2x=9

=>x=9/2

b)3x-6=x^2-16

=>3x-6-x^2+16=0

=>3x+10-x^2=0

=>-x^2+3x+10=0

=>-(x^2-3x-10)=0

=>-(x^2-5x+2x-10)=0

=>-[(x-5)+2(x-5)]=0

=>-[(x-5)(x+2)]=0

=>-(x-5)(x+2)=0

=> 2 TH:

*x-5=0

=>x=0+5

=>x=5

*x+2=0

=>x=0-2

=>x=-2

vậy x=5 hay x=-2

c)(2x-3)-2-49=0

=>(2x-3)-2=49

=>(2x-3)=49+2

=>2x-3=51

=>2x=51+3

=>2x=54

=>x=54/2=27

vậy x=27