Bài làm:

a, 1-4x²

=1-(2x)²

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

b, 8-27x³

=2³-(3x)³

=(2-3x).(4+6x+9x²)

Các câu còn lại bạn dùng hằng đẳng thức là phân tích được ra thôi

1 - 4x^2 

= 1^2 - ( 2x )^2 

= ( 1 - 2x ) ( 1 + 2x ) 

8 - 27x^ 3 

= 2^3 - ( 3x )^3 

= ( 2 - 3x ) [ 2^2 + 2 * 3x + ( 3x )^2 ]

= ( 2 - 3x ) ( 4 + 6x + 9x^2 ) 

= ( 2 - 3x ) ( 9x^2 + 6x + 4 ) 

27 + 27x + 9x^2 + x^3 

= x^3 + 9x^2 + 27x + 27 

= x^3 + 3x^2 + 6x^2 + 18x + 9x + 27 

= x^2 ( x + 3 ) + 6x ( x + 3 ) + 9 ( x + 3 ) 

= ( x + 3 ) ( x^2 + 6x + 9 ) 

= ( x + 3 ) ( x + 3 )^2 

= ( x + 3 )^3 

x^2 + 4x - 5 

= x^2 - x + 5x - 5 

= x ( x - 1 ) + 5 ( x - 1 ) 

= ( x + 1 ) ( x - 5 ) 

\(\left(x-1\right)^2-25\)

\(=x^2-2x+1-25\)

\(=x^2-2x-24\)

\(=x^2-6x+4x-24\)

\(=x.\left(x-6\right)+4.\left(x-6\right)\)

\(=\left(x+4\right).\left(x-6\right)\)

a, \(1-2y+y^2=\left(y+1\right)^2=\left(y+1\right)\left(y+1\right)\)

b, \(\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

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

d,  \(8-27x^3=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

1. a) \(8x^3-32x=8x\left(x^2-4\right)=8x\left(x-4\right)\left(x+4\right)\)

b) \(y^3+64+\left(y+4\right)\left(y-16\right)=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)

\(=\left(y+4\right)\left(y^2-4y+16\right)+\left(y+4\right)\left(y-16\right)=\left(y+4\right)\left(y^2-4y+16+y-16\right)\)

\(=\left(y-4\right)\left(y^2-3y\right)=\left(y-4\right)y\left(y-3\right)\)

2) a)

\(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow x\left(2x+3\right)\left(2x-3\right)=0\)

<=> x=0 hoặc 2x+3=0 hoặc 2x-3=0

<=> x=0 hoặc x=-3/2 hoặc x=3/2

b) \(A=x^3-9x^2+27x-27=x^3-3.x^2.3+3.x.3^2-3^3=\left(x-3\right)^3\)

Tại x=203

A=(203-3)³=200³

Bài 1 :

a) \(8x^3-32x\)

\(=8x\left(x^2-4\right)\)

\(=8x\left(x-2\right)\left(x+2\right)\)

b) \(y^3+64+\left(y+4\right)\left(y-16\right)\)

\(=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)

\(=\left(y+4\right)\left(y^2-4y+16\right)+\left(y+4\right)\left(y-16\right)\)

\(=\left(y+4\right)\left(y^2-4x+16+y-16\right)\)

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

Bài 2 :

a) \(4x^3-9x=0\)

\(x\left(4x^2-9\right)=0\)

\(x\left[\left(2x\right)^2-3^2\right]=0\)

\(x\left(2x-3\right)\left(2x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\2x-3=0\\2x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\\x=\frac{-3}{2}\end{cases}}}\)

P.s: ở trên dùng ngoặc vuông nhé

b) \(A=x^3-9x^2+27x-27\)

\(A=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)

\(A=\left(x-3\right)^3\)

Thay x = 203 vào biểu thức ta có :

\(A=\left(203-3\right)^3\)

\(A=200^3\)

\(A=8000000\)

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

\(8-27x^3\)

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

\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)

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

c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

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

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

\(=-x^3+3x^2+6x^2-18x-9x+27\)

\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)\)

\(=-\left(x-3\right)\left(x^2-6x+9\right)\)

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

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

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

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

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

2: Ta có: \(9x^2+6x-4y^2-4y\)

\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)

\(=\left(3x-2y\right)\left(3x+2y+2\right)\)

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

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

\(=x^3+3x^2+6x^2+18x+9x+27\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+9\right)\)

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

Nếu nhìn kĩ thì bạn sẽ thấy đây là hằng đẳng thức nhé !

\(x^3+9x^2+27x+27=x^3+3.3.x^2+3.3^2.x+3^3\)

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