=(x-5)(x-4)

\(=x^2-4x-5x+20=\left(x-4\right)\left(x-5\right)\)

\(x^2-9x+20=\left(x^2-4x\right)-\left(5x-20\right)=x\left(x-4\right)-5\left(x-4\right)=\left(x-4\right)\left(x-5\right)\)

\(9x^2-4\left(x-1\right)^2\)

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

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

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

Trí làm đúng rồi nha

\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)

\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)

\(-x^3+9x^2-27x+27=\left(3-x\right)^3\)

\(-x^3+9x^2-27x+27\)

\(=-x^3+3x^2+6x^2-18x-9x+27\)

\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)\)

\(=-\left(x-3\right)\left(x^2-6x+9\right)\)

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

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

       \(x^3-9x^2+14x\)

= \(x^3-7x^2-2x^2+14x\)

= \(x^2.\left(x-7\right)-2x.\left(x-7\right)\)

= \(\left(x-7\right).\left(x^2-2x\right)\)

= \(\left(x-7\right).\left(x-2\right).x\)

\(x^3-9x^2+14x\)

\(=x\left(x^2-9x+14\right)\)

\(=x\left(x^2-7x-2x+14\right)\)

\(=x\left[x\left(x-7\right)-2\left(x-7\right)\right]\)

\(=x\left(x-2\right)\left(x-7\right)\)

khó quá tui ko biết làm..

k cho tui nha

thanks

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)

\(9x^2-12xy+4y^2\)

\(=\left(3x\right)^2-2.3x.2y+\left(2y\right)^2\)

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

\(\left(x^2-3x+2\right)\left(x^2-9x+20\right)-40=\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-40\)

\(=\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40\)
Đặt \(t=x^2-6x+5\) thì ta có \(t\left(t+3\right)-40=t^2+3t-40=\left(t+8\right)\left(t-5\right)\)

Suy ra \(\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40=\left(x^2-6x+13\right)\left(x^2-6x\right)=x\left(x-6\right)\left(x^2-6x+13\right)\)