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

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

\(8x\left(x^2-9\right)=0\Rightarrow8x\left(x-3\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)

\(x^2-8x-9\)

\(=x^2-9x+x-9\)

\(=x\left(x-9\right)+\left(x-9\right)\)

\(=\left(x-9\right)\left(x+1\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

x²-8x + 16 - 4 = ( x - 4 )²-2²= ( x -4-2 ) . ( x-4+2 ) = ( x - 6 ) .( x -2 )

a) Ta có: \(8x+4x^2-12xy\)

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

b) Ta có: \(5x^3-10x^2+5x\)

\(=5x\left(x^2-2x+1\right)\)

\(=5x\left(x-1\right)^2\)

c) Ta có: \(x^3+x^2y-xy^2-y^3\)

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

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

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

d) Ta có: \(x^2-8x-9\)

\(=x^2-9x+x-9\)

\(=\left(x-9\right)\left(x+1\right)\)

a. `8x+4x^2-12xy=4x(2+x-3y)`

b) `5x^3-10x^2+5x=5x(x^2-2x+1)`

c) `x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y)=(x+y)(x^2-y^2)=(x+y)^2 (x-y)`

d) `x^2-8x-9=(x^2-2.x.4+4^2)-25=(x-4)^2-5^2=(x+1)(x-9)`

( x² + 8x + 7 ) ( x² + 8x + 15 ) + 15

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

đặt t = x² + 8x + 7

có : t ( t + 8 ) + 15 

= t² + 8t + 15

= t² + 5t + 3t + 15

= ( t² + 5t ) + ( 3t + 15 )

= t ( t + 5 ) + 3 ( t + 5 )

= ( t + 3 ) ( t + 5 )

mà t = x² + 8x + 7

⇒ ( x² + 8x + 10 ) ( x² + 8x + 12 )

C1

Ta có \(x^2-8x+12\)

\(=x^2-6x-2x+12\)

\(=x\left(x-6\right)-2\left(x-6\right)\)

\(=\left(x-6\right)\left(x-2\right)\)

C2 

ta có \(x^2-8x+12\)

\(=x^2-8x+16-4\)

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

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

\(=\left(x-6\right)\left(x-2\right)\)

\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)

=(x-2)(x-6)