\(x^3+9x^2+26x+24\)

\(=x^3+3x^2+6x^2+18x+8x+24\)

\(=\left(x^3+3x^2\right)+\left(6x^2+18x\right)+\left(8x+24\right)\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+8\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+8\right)\)

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

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

\(=\left(x+3\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)

\(=\left(x+3\right)\left(x+2\right)\left(x+4\right)\)

Ta có : 15x³ + 29x² - 8x - 12

= 15x³ + 30x² - x² - 8x - 12

= 15x(x + 2) - (8x + 16) - (x² - 4)

= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)

= (x + 2)(15x - 8 - x + 2)

= (x + 2) (14x - 6) 

 click zô nha >_<                 

Ta có : 15x³ + 29x² - 8x - 12

= 15x³ + 30x² - x² - 8x - 12

= 15x(x + 2) - (8x + 16) - (x² - 4)

= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)

= (x + 2)(15x - 8 - x + 2)

= (x + 2) (14x - 6)

(x+2)(14x-6)

\(4x^2-4x-35\) \(=\left(2x\right)^2-2.2x.1+1-36\)

                                     \(=\left(2x-1\right)^2-6^2\)

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

\(18x^2-5x-2\) \(=\left(x-\frac{1}{2}\right)\left(x+\frac{2}{9}\right)\)

\(8x^3-26x^2+13x+5=\)  \(8x^3-8x^2-18x^2+18x-5x+5\)

                                                \(=8x^2\left(x-1\right)-18x\left(x-1\right)-5\left(x-1\right)\)

                                                \(=\)  \(\left(8x^2-18x-5\right)\left(x-1\right)\)

                                                \(=\left(x-\frac{5}{2}\right)\left(x+\frac{1}{4}\right)\)\(\left(x-1\right)\)

a.\(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

b.\(x^3-6x^2+12x-8=\left(x-2\right)^3\)

c.\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

\(2x-1^3+8\)

\(=2x-9\)

\(=\left(\sqrt{2x}\right)^2-3^2\)

\(=\left(\sqrt{2x}-3\right)\left(\sqrt{2x}+3\right)\)

_________

\(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

_______________

\(8x^3-12x^2+6x-2\)

\(=8x^3-12x^2+6x-1-1\)

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

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

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

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

________

\(9x^3-12x^2+6x-1\)

\(=x^3+8x^3-12x^2+6x-1\)

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

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

\(=\left(3x-1\right)\left(3x^2-5x+1\right)\)

b: 8x^3-12x^2+6x-1

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

=(2x-1)^3

c: =(8x^3-12x^2+6x-1)-1

=(2x-1)^3-1

=(2x-1-1)[(2x-1)^2+2x-1+1]

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

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

8x³ - 12x² + 6x - 1

= (2x)³ - 3.(2x)².1 + 3.2x.1 - 1³

= (2x - 1)³

--------------------

8x³ - 12x² + 6x - 2

= 8x³ - 12x² + 6x - 1 - 1

= (2x)³ - 3.(2x)².1 + 3.(2x).1 - 1³ - 1³

= (2x - 1)³ - 1³

= (2x - 1 - 1)[(2x - 1)² + (2x - 1).1 + 1]

= (2x - 2)(4x² - 4x + 1 + 2x - 1 + 1)

= 2(x - 1)(4x² - 2x + 1)

--------------------

9x³ - 12x² + 6x - 1

= x³ + 8x³ - 12x² + 6x - 1

= x³ + (2x)³ - 3.(2x)² + 3.2x.1² - 1³

= x³ + (2x - 1)³

= (x + 2x - 1)[x² - x.(2x - 1) + (2x - 1)²]

= (3x - 1)(x² - 2x² + x + 4x² - 4x + 1)

= (3x - 1)(3x² - 3x + 1)

Nếu bạn muốn có lời giải thì ít thôi @.@

Katherine Lilly Filbert nói rất đúng câu hỏi nhiều như vậy ai mà trả lời đc hết cơ chứ

Câu c đề sai thì phải bạn ạ.

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

b) \(-x^3+12x^2-48x+64=\left(4-x\right)^3\)

c) \(27+27x+9x^2=9\left(x^2+3x+3\right)\)