\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

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

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

\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)

Đặt x² + 4x + 8 = A. Ta sẽ được:

A² + 3xA + 2x² 

= A² - xA - 2xA + 2x²

= A(A-x) - 2x(A-x)

= (A-x)(A-2x)

= (x²+3x+8)(x²+2x+8)

6, (x^2 +1) -4x^2 = x^2 + 1 - 4x^2 = 1 - (4x^2 - x^2) = 1 - 3x^2 = (1-\(\sqrt{3}\)x)(1+\(\sqrt{3}\)x)

7, x^2 - 4x -5 = x^2 - 2.x.2 + 4 - 9 = (x^2 - 2.x.2 +4) - 3^2 = (x-2)^2 - 3^2 = (x-2-3)(x-2+3) = (x-5)(x+1)

8, x^5 - 3x^4 + 3x^3 - x^2 = x^2(x^3 -3x^2 + 3x -1) = x^2(x-1)^3

\(x^5-3x^4+3x^3-x^2\)

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

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

hk tốt

^^

đặt y=x²+4x+8 ta được

y²+3xy+2x²=y²+xy+2xy+2x²=y(y+x)+2x(y+x)

=(y+x)(y+2x)

thay y=x²+4x+8 ta được

(x²+5x+8)(x²+7x+8)

=(x^2+4x+8)²+2x(x^2+4x+8)+(x^2+4x+8)+2x^2

=(x^2+5x+8)(x^2+6x+8)

(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2

dat x^2+4x+8=y

ta co:y^2+3xy+2x^2

=y^2+xy+2xy+2x^2

=y(y+x)+2x(y+x)

=(y+2x)(y+x)

=(x^2+4x+8+2x)(x^2+4x+8+x)

=(x^2+6x+8)(x^2+5x+8)

KL:......................

\(a,=x\left(y-3\right)+y\left(y-3\right)=\left(x+y\right)\left(y-3\right)\\ b,=\left(x+2\right)^2-16y^2=\left(x+4y+2\right)\left(x-4y+2\right)\)

\(a,xy-3x+y^2-3y=\left(xy-3x\right)+\left(y^2-3y\right)=x\left(y-3\right)+y\left(y-3\right)=\left(x+y\right)\left(y-3\right)\\ b,x^2-16y^2+4x+4=\left(x^2+4x+4\right)-16y^2=\left(x+2\right)^2-\left(4y\right)^2=\left(x-4y+2\right)\left(x+4y+2\right)\)