a: \(1-\dfrac{x^3}{8}=\left(1-\dfrac{1}{2}x\right)\left(1+\dfrac{1}{2}x+\dfrac{1}{4}x^2\right)\)

b: \(27x^3+1=\left(3x+1\right)\left(9x^2-3x+1\right)\)

c: \(64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

A) \(x^3+27\)

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

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

B) \(x^3-\dfrac{1}{8}\)

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

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

C) \(8x^3+y^3\)

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

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

D) \(8x^3-27y^3\)

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

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

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

b)\(\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)

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

d)\(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)

\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

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

b) \(64x^3+0,001=\left(4x\right)^3+\left(\dfrac{1}{10}\right)^3=\left(4x+\dfrac{1}{10}\right)\left(16x^2-\dfrac{4x}{10}+\dfrac{1}{100}\right)\)

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

b/ \(64x^3+0,001=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\)

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

d/ \(\dfrac{x^3}{125}-\dfrac{y^3}{27}=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)

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

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

c: \(x^3+125=\left(x+5\right)\left(x^2-5x+25\right)\)

d: \(x^3-27y^3=\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

a) 8x³ - 1

= (2x)³ - 1³

= (2x - 1)(4x² + 2x + 1)

b) x³ + 8y³

= x³ + (2y)³

= (x + 2y)(x² + 2xy + 4y²)

c) x³ + 125

= x³ + 5³

= (x + 5)(x² - 5x + 25)

d) x³ - 27y³

= x³ - (3y)³

= (x - 3y)(x² + 3xy + 9y²)

 Chúc bạn học tốt

a. 8x³ - 1

⇔ (2x)³ - 1³

⇔ (2x - 1)³.

b. x³ + 8y³

⇔ x³ + (2y)³

⇔ (x + 2y)³

c. x³ + 125

⇔ x³ + 5³

⇔ (x + 5)³

d. x³ - 27y³

⇔ x³ - (3y)³

⇔ (x - 3y)³

a)8x3 + * + * + 27y3 = (* + *)3

=>A=(2x+3y)^3

b) (2x+1)^3

c)(x-2y)^3

d)(3x-2)(3x+2)

e)(3x-1)(9x^2+3x+1)

f)....................

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

7: \(\left(2x+1\right)^2=4x^2+4x+1\)

8: \(\left(2x-1\right)^2=4x^2-4x+1\)

9: \(9-16x^2=\left(3-4x\right)\left(3+4x\right)\)

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

\(\dfrac{9}{16}-\dfrac{1}{4}x^2=\left(\dfrac{3}{4}-\dfrac{1}{2}x\right)\left(\dfrac{3}{4}+\dfrac{1}{2}x\right)\)