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: \(3x^4-4x^3+1\)

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

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

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

b: \(x^3-19x-30\)

\(=x^3-4x-15x-30\)

\(=x\left(x-2\right)\left(x+2\right)-15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x-15\right)\)

\(=\left(x+2\right)\cdot\left(x-5\right)\left(x+3\right)\)

Em xem gõ lại đề cho đúng nha

`#\text {Kr.Ryo}`

`a)`

`4x^2 - 4x + 1`

`= (2x)^2 - 2*2x*1 + 1^2`

`= (2x - 1)^2`

`b)`

Xem lại đề

`c)`

`2x^2 + 7x + 5`

`= 2x^2 + 2x + 5x + 5`

`= (2x^2 + 2x) + (5x + 5)`

`= 2x(x + 1) + 5(x + 1)`

`= (2x + 5)(x + 1)`

`d)`

`x^2 - 6xy - 25z^2 + 9y^2`

`= (x^2 - 6xy + 9y^2) - 25z^2`

`= [ (x)^2 - 2*x*3y + (3y)^2] - (5z)^2`

`= (x + 3y)^2 - (5z)^2`

`= (x + 3y - 5z)(x + 3y + 5z)`

\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)

a) 

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

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

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

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

a) \(x^2+5x+4==x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)

b) \(3x^2+4x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(x-1\right)\left(3x+7\right)\)

c) \(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

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

b) (x-1)(3x+7)

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

a) x²+5x+4 = x(x+4)+(x+4) = (x+4)(x+1)

b) 3x²+4x-7 = 3x(x-1)+7(x-1) = (x-1)(3x+7)

c) x²+7x+12 = x(x+4)+3(x+4) = (x+3)(x+4)