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

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

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

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

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

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

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

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

(x2+2x)+9x2+18x+20

=(x2+2x)+9(x2+2x)+20

Đặt t=x2+2x đc:

t+9t+20=10t+20=10(t+2)

Thay t=x2+2x vào đc:

10(x2+2x+2)

(x²+2x)²+9x²+18x+20

=(x²+2x)²+9(x²+2x)+20

Đặt t=x²+2x ta được:

t²+9t+20=t²+4t+5t+20

=t.(t+4)+5.(t+4)

=(t+4)(t+5)

thay t=x²+2x ta được:

(x²+2x+4)(x²+2x+5)

Vậy (x²+2x)²+9x²+18x+20=(x²+2x+4)(x²+2x+5)

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

\(=x\cdot x-x\cdot9\)

=x(x-9)

(x² + 2x + 4) (x² + 2x + 5)

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

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

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

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

Trí làm đúng rồi nha

\(=x^2-4x-5x+20\) 

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

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

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)