a: =(x+y)^3+z^3-3xy(x+y)-3xyz

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)

\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

=(x^2+x+5)(x^2+x-2)

=(x^2+x+5)(x+2)(x-1)

d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2

=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c

=b^2(c-a)+b(c^2-a^2)+ac(c-a)

=(c-a)(b^2+ac)+b(c-a)(c+a)

=(c-a)(b^2+ac+bc+ba)

=(c-a)[b^2+bc+ac+ab]

=(c-a)[b(b+c)+a(b+c)]

=(c-a)(b+c)(b+a)

\(a,=\left(x-2\right)\left(15x-7y\right)\\ b,=x\left(x-11\right)\left(2x-1\right)\\ c,=2x\left(x-3\right)\left(2+3y\right)\\ d,=\left(x-y\right)\left(x-7y\right)\\ e,=\left(x-3\right)\left(4x-12-2x\right)\\ =\left(x-3\right)\left(2x-12\right)=2\left(x-6\right)\left(x-3\right)\)

\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)

1.

a) \(2x^4-4x^3+2x^2\)

\(=2x^2\left(x^2-2x+1\right)\)

\(=2x^2\left(x-1\right)^2\)

b) \(2x^2-2xy+5x-5y\)

\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)

\(=2x\left(x-y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\cdot\left(2x+5\right)\)

2 . 

a,

\(4x\left(x-3\right)-x+3=0\)

⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)

⇒\(\left(x-3\right)\left(4x-1\right)=0\)

⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)

vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)

b, 

\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)

⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0

⇒\(\left(x-4\right)\left(3x-2\right)=0\)

⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)

vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)

ban tích cho mk vs nha

a) (x - 1)(x + l)(x - 2)(x - 4).      b) (x - 2)( x 2  + 4).

c) 2y(3 x 2   +   y 2 ).                          d) 2(x + y + z) ( a   -   b ) 2 .

a. \(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)

\(=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)

\(=\left[\left(x-3\right)^2-1\right]\left(x^2-1\right)\)

\(=\left(x-3+1\right)\left(x-3-1\right)\left(x+1\right)\left(x-1\right)\)

\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\left(x-1\right)\)

b. \(x^3-2x^2+4x-8\)

\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)

\(=x\left(x^2+4\right)-2\left(x^2+4\right)\)

\(=\left(x-2\right)\left(x^2+4\right)\)

c. \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=6x^2y+2y^3\)

\(=2y\left(3x^2+y^2\right)\)

d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)

\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)

\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)

\(=2\left(a-b\right)^2\left(x+y+z\right)\)

a: Ta có: \(x^2-6x+9-y^2\)

\(=\left(x-3\right)^2-y^2\)

\(=\left(x-y-3\right)\left(x+y-3\right)\)

b: Ta có: \(x^3+4x^2+4x\)

\(=x\left(x^2+4x+4\right)\)

\(=x\left(x+2\right)^2\)

c: Ta có: \(4xy-4x^2-y^2+9\)

\(=-\left(4x^2-4xy+y^2-9\right)\)

\(=-\left(2x-y-3\right)\left(2x-y+3\right)\)

a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1=\left(x^2+3x+1\right)^2\)

b) \(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)=25\Leftrightarrow1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)-25=0\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)\left(1+xy\right)+\left(1+xy\right)^2-25=0\Leftrightarrow\left(x+y+1+xy\right)^2-25=0\Leftrightarrow\left(x+y+xy-24\right)\left(x+y+xy+26\right)=0\)

a: Ta có: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)

\(=\left(x^2+3x+1\right)^2\)