\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)

\(x^3-6x^2-24x+144=\left(x^3-6x^2\right)-\left(24x-144\right)=x^2\left(x-6\right)-24\left(x-6\right)=\left(x-6\right)\left(x^2-24\right)\)

b: \(-5x^2+16x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2-x+2021x-2021=0\\ \Leftrightarrow\left(x-1\right)\left(x+2021\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2021\end{matrix}\right.\\ b,\Leftrightarrow-5x^2+15x+x-3=0\\ \Leftrightarrow\left(x-3\right)\left(1-5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

= x^2 - 7xy + 10y^2
= x^2 - 2xy -5xy + 10y^2
= x( x - 2y ) - 5y( x - 2y )
= ( x - 2y )( x - 5y )

x² – 7xy + 10y2 

= x² – 2xy – 5xy + 10y²

= x(x – 2y) – 5y(x – 2y)

= (x – 2y)(x – 5y)

27a²b²-18ab+3=3(9a²b²-6ab+1)=3(3ab-1)²

$=x^2-2xy-5xy+10y^2\\=x(x-2y)-5y(x-2y)\\=(x-5y)(x-2y)$

x^2 - 7xy + 10y^2 
= (x^2 - 2xy) - (5xy - 10y^2) 
= x(x - 2y) - 5y( x - 2y) 
= (x - 5y)(x - 2y) 

mik mới học lớp 7 thôi!

Ta có : x³ - 3x² + 1 - 3x

= x³ - 2x² - x - (x² - 2x - 1) 

= x(x² - 2x - 1) - (x² - 2x - 1) 

= (x² - 2x - 1) (x - 1) 

\(a,x^2\left(x-2\right)-4x+8\\ =\left(x^2-4\right)\left(x-2\right)\\ =\left(x-2\right)^2\left(x+2\right)\\ b,x^2+7xy+10y^2\\ =x^2+2xy+5xy+10y^2\\ =x\left(x+2y\right)+5y\left(x+2y\right)\\ =\left(x+5y\right)\left(x+2y\right)\)