\(a,x^3+6x^2y+12xy^2+8y^3\\ =x^3+3.2x^2+3.2^2.x+\left(2y\right)^3\\ =\left(x+2y\right)^3\)

\(b,x^3-3x^2+3x-1\\ =x^3-3x^2.1+3x.1^2-1^3\\ =\left(x-1\right)^3\)

a) \(x^3+6x^2y+12xy^2+8y^3\)

\(=x^3+3\cdot x^2\cdot2y+2\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)

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

b) \(x^3-3x^2+3x-1\)

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

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

Giải:

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

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

\(=-\left(x^3-3.x^2.1+3.x.1^2-1^3\right)\)

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

Vậy ...

Chúc bạn học tốt!

–x3 + 3x2 – 3x + 1

= (–x)3 + 3.(–x)2.1 + 3.(–x).1 + 13

= (–x + 1)3 (Áp dụng HĐT (4) với A = –x và B = 1)

a) \(A=8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)

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

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

d)  \(D=27+27y^2+9y^4+y^6=\left(3+y^2\right)^3\)

a) ta có : \(-x^3+3x^2-3x+1=1-3.1^2.x+3.1.x^2-x^3=\left(1-x\right)^3\)

b) ta có : \(64-48x+12x^2-x^3=4^3-3.4^2x+3.4.x^2-x^3=\left(4-x\right)^3\)

Ta có: a/ -x³+3x²-3x+1 = -(x³-3x²+3x-1)

= -(x-1)³

b/ 64-48x+12x²-x³ = 4³-3.4².x+3.a.x²-x³

= (4-x)³

a,(1-x)^3

b,(4-x)^3

Bài giải:

a) – x³ + 3x²– 3x + 1 = 1 – 3 . 1² . x + 3 . 1 . x² – x³

= (1 – x)³

b) 8 – 12x + 6x² – x³ = 2³ – 3 . 2². x + 3 . 2 . x² – x³

= (2 – x)³

a) \(-x^3+3x^2-3x+1=1-3x+3x^2-x3\)

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

b) \(8-12x+6x^2-x^3\) \(=2^3-3.2^2.x+3.2.x^2-x^3\)

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

a) -x³+3x²-3x+1=1-3x+3x²-x³

=(1-x)³

b) 8-12x+6x²-x³=(2-x)³

x³-3x²+3x-1

= (x-1)³

a) -x³ + 3x² - 3x + 1

= -(x³ - 3x² + 3x - 1)

=-(x - 1)³

b) 8 - 12x + 6x² - x³

= 2³ - 3.2².x + 3.2.x² - x³

= (2 - x)³

1, \(x^2+2xy+y^2=\left(x+y\right)^2\)

2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)

3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)

4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)

5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

1: =(x+y)^2

2: =(2x+3)^2

3: =(x+5/2)^2

4: =(4x-1)^2

5: =(x+1/2)^2

6: =(x-3/2)^2

7: =(x+1)^3

8: =(1/2x+1)^2

9: =(3y-1/3)^3

10: =(2x+y)^3

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

7, \(x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\)

8, \(\dfrac{x^2}{4}+x+1=\left(\dfrac{x}{2}\right)^2+2\cdot\dfrac{x}{2}\cdot1+1^2=\left(\dfrac{x}{2}+1\right)^2\)

9, \(27y^3-9y^2+y-\dfrac{1}{27}=\left(3y\right)^3-3\cdot\left(3y\right)^2\cdot\dfrac{1}{3}+3\cdot3y\cdot\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3=\left(3y-\dfrac{1}{3}\right)^3\)

10, \(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3=\left(2x+y\right)^3\)

a ) Ta có : -x³ + 3x² - 3x + 1

= 1 - 3x + 3x² - x³

= (1 - x)³ 

b) Ta có : 8 - 12x + 6x² - x³

= 2³ - 3.2².x + 3.2.x² - x³

= (2 - x)³

a, -x³ + 3x² - 3x + 1

   = -x³ + 3.x².1 - 3.x.1² + 1³ 

   = ( -x + 1 )³