Ta có:

\(x^2+3x-18\)

\(\Leftrightarrow x^2-3x+6x-18\)

\(\Leftrightarrow\left(x^2-3x\right)+\left(6x-18\right)\)

\(\Leftrightarrow x\left(x-3\right)+6\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x+6\right)\)

\(x^2+3x-18\)

\(\Leftrightarrow x\left(x+3\right)=18\)

\(\Leftrightarrow\orbr{\begin{cases}x=18\\x+3=18\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=18\\x=15\end{cases}}\)

vay \(\orbr{\begin{cases}x=18\\x=15\end{cases}}\)

thank bn

\(=x^2-4x-5x+20\) 

\(=x\left(x-4\right)-5\left(x-4\right)\) 

\(=\left(x-4\right)\left(x-5\right)\)     

x² - 9x + 20

= x² - 4x - 5x + 20

= x( x - 4 ) - 5( x - 4 )

= ( x - 4 )( x - 5 )

x²-9x+20=(x²-4x)-(5x-20)=x(x-4)-5(x-4)=(x-4)(x-5)

x² + 3x + 6x + 18

= x(x + 3) + 6(x + 3)

= (x + 6)(x + 3)

dễ ẹc!!!!!!!!

\(x^2+9x+18\)

\(=x^2+6x+3x+18\)

\(=x\left(x+6\right)+3\left(x+6\right)\)

\(=\left(x+3\right)\left(x+6\right)\)

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

=x(x²+2x+1)-3²

=x(x+1)²-3²

=x(x+1-3)(x+1+3)

a: \(9x^3y^2+3x^2y^2\)

\(=3x^2y^2\cdot3x+3x^2y^2\cdot1\)

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

b: \(x^2-2x+1-y^2\)

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

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

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

c: \(x^4+x^3-4x^2+x+1\)

\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)

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

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

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

a³ + b³ + c³ - 3abc = (a + b)³ + c³ - 3abc - 3ab(a + b)

= (a + b + c)(a² + b² + 2ab - ac - bc + c²) - 3ab(a + b + c)

= (a + b + c)(a² + b² + c² - ab - ac - bc)

\(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)

\(=2x^4+6x^3+9x^2+6x+2\)(bạn nhân phá ngoặc rồi thu gọn nhé)

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

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

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