a) 

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

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

\(a,=2x^2-6x+9x-27=\left(x-3\right)\left(2x+9\right)\\ b,=x^2-7x+\dfrac{49}{4}-\dfrac{73}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{73}{4}=\left(x-\dfrac{7}{2}-\dfrac{\sqrt{73}}{2}\right)\left(x-\dfrac{7}{2}+\dfrac{\sqrt{73}}{2}\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ d,=x^2-2x-8x+16=\left(x-2\right)\left(x-8\right)\\ e,=x^2-3x-5x+15=\left(x-3\right)\left(x-5\right)\\ g,=x^2+2x+4x+8=\left(x+2\right)\left(x+4\right)\)

Lời giải:
a.

$x^2-7x+6=(x^2-x)-(6x-6)=x(x-1)-6(x-1)=(x-1)(x-6)$

b.

$x-3\sqrt{3}x-12\sqrt{3}$ không phân tích được thành nhân tử

c.

$x^2+4x-2$ không phân tích được thành nhân tử với các hệ số nguyên.

a)\(x^2+10x+25-y^2\)

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

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

b)\(5x^3-7x^2+10x-14\)

\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)

\(=\left(x^2+2\right)\left(5x-7\right)\)

c)\(-5y^2+30y-45\)

\(=-5\left(y^2-6y+9\right)\)

\(=-5\left(y-3\right)^2\)

e)\(4xy^2-8xyz+4xz^2\)

\(=4x\left(y^2-2yz+z^2\right)\)

\(=4x\left(y-z\right)^2\)

f)\(x^2+7x+10\)

\(=x^2+5x+2x+10\)

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

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

k)\(2x^7+6x^6+6x^5-2x^4\)

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

a)\(x^2+10x+25-y^2\)

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

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

b)\(5x^3-7x^2+10x-14\)

\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)

\(=\left(5x-7\right)\left(x^2+2\right)\)

c)\(-5y^2+30y-45\)

\(=-5\left(y^2-6y+9\right)\)

\(=-5\left(y-3\right)^2\)

e)\(4xy^2-8xyz+4xz^2\)

\(=4x\left(y^2-2yz+z^2\right)\)

\(=4x\left(y-z\right)^2\)

f)\(x^2+7x+10\)

\(=x^2+5x+2x+10\)

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

k)\(2x^7+6x^6+6x^5-2x^4\)

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

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

làm lại

6x³ + 5x² - 7x - 4

= (6x³ + 5x - 7x) - 4

= x (6x² - 5 - 7) - 2²

= x (6x² - 12) - 2²

= x [6 (x² - 2)] - 2²

= x [6 (x² - \(\sqrt{2}^2\))] - 2²

= x [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))] - 2²

= (x - 2²) [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))

b) 2x³ - x² + x - 2

= (2x³ - x² - x) - 2

= x (2x² - x - 1) - 2

= (x - 2) (2x² - x - 1)

(mik ko biet dug ko, neu sai mog bn thog cam)

\(a,=5x^2-5x+3x-3=\left(x-1\right)\left(5x+3\right)\\ b,=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\\ c,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ d,=7x^2-7x+x-1=\left(x-1\right)\left(7x+1\right)\)

c: =(x+5)(x-3)

d: =(x-1)(7x+1)

\(a,5x^2-2x-3=\left(5x^2-5x\right)+\left(3x-3\right)=5x\left(x-1\right)+3\left(x-1\right)=\left(x-1\right)\left(5x+3\right)\\ b,2x^2-3x-5=\left(2x^2+2x\right)-\left(5x+5\right)=2x\left(x+1\right)-5\left(x+1\right)=\left(x+1\right)\left(2x-5\right)\\ c,x^2+2x-15=\left(x^2-3x\right)+\left(5x-15\right)=x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(x+5\right)\\ d,7x^2-6x-1=\left(7x^2-7x\right)+\left(x-1\right)=7x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(7x+1\right)\)

a, x^2 + 5x +4

= x^2 + 1x + 4x + 4

= (x^2 + 1x) + (4x + 4)

= x ( x + 1 ) + 4 ( x + 1 )

= (x + 1) (x + 4)

b, x^2 - 6x + 5

= x^2 - 1x - 5x + 5

= (x^2 - 1x) - (5x - 5)

= x (x - 1) - 5 (x - 1)

= (x - 1) (x - 5)

c, x^2 + 7x + 12

= x^2 + 3x + 4x + 12 

= (x^2 + 3x) + (4x + 12)

= x (x + 3) + 4 (x + 3)

= (x + 3) (x + 4)

d, 2x^2 - 5x + 3

= 2^x2 - 2x - 3x + 3

= 2x (x - 1) - 3 (x - 1)

= (x-1) (2x - 3)

e, 7x  - 3x^2 - 4

= 3x + 4x - 3x^2 - 4

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

= 3x (1 - x) + 4 (x - 1)

= 3x (1-x) - 4 (1 - x)

= (1 - x) (3x - 4)

f, x^2 - 10x + 16

= x^2 - 2x - 8x + 16

= (x^2 - 2x) - (8x - 16)

= x (x - 2) - 8 (x - 2)

= (x - 2) (x - 8)

a, (x+1)(x+4)

b,(x-5)(x-1)

c,(x+3)(x+4)

d,(2x-3)(x-1)

e,(-3x+4)(x-1)

f, (x-8)(x-2)

\(a,=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)