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:

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

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

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

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

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

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

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

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

cái này dễ mà

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

= (2x-1)^3

\(a,x^2+6x+9\)

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

\(b,10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

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

\(c,8x^3-\frac{1}{8}\)

\(=8x^3-\left(\frac{1}{2}\right)^3\)

\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)

\(d,8x^3+12x^2+6xy^2+y^3\)

\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)

hok tốt!

ai tra loi dung minh cho

Điệp viên 007 sai c

c, \(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

9x²-12x+4

=(3x)²-2.3x.2+2²

=(3x-2)²

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

a: Sửa đề: x^3-x^2+5x-5

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

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

b: x^3+4x^2+x-6

=x^3-x^2+5x^2-5x+6x-6

=(x-1)(x^2+5x+6)

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

c: \(=\left(x+2\right)^3+y^3\)

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