(4a⁴-12a²+1)=(4a⁴-8a²+1)-4a²=(2a²-1)²-4a²=(2a²-2a-1)(2a²+2a-1)

tu de bai suy ra: 3x(x^3+4x+4)=3x(x+2)^2

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

Lời giải:

$x^3-4x^2-12x+27$

$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$

$=x^2(x+3)-7x(x+3)+9(x+3)$

$=(x+3)(x^2-7x+9)$

a) = a( a^2-1 )

= a(a-1)(a+1)

b) =16x^3- 16x^2 +4x^2-4x+7x -7

=16x^2(x-1) +4x(x-1) +7(x-1)

=(x-1)(16x^2 +4x+7)

bạn có thể viết rõ ràng hơn đc k ? mình k hiểu đề bài cho lắm!!!

a) a^3-a

b)16x^3-12x^2+3x-7

phân tích thành nhân tử

cái này dễ mà

= (2x)^3-3(2x)^2*1+2*3x*1^2-1^3

= (2x-1)^3

Câu a :

\(\dfrac{3x^2-12x+12}{x^4-8x}\)

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

\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

Câu b :

\(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

\(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=\left(4y\right)^2-\left(2x+3\right)^2\)\(=\left[4y-\left(2x+3\right)\right]\left(4y+2x+3\right)=\left(4y-2x-3\right)\left(4y+2x+3\right)\)

\(x^3-4x^2+12x-27\)

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

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

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