\(\dfrac{27x^3}{y^3}+\dfrac{8y^3}{125}\left(y\ne0\right)\\ =\left(\dfrac{3x}{y}\right)^3+\left(\dfrac{2y}{5}\right)^3\\ =\left(\dfrac{3x}{y}+\dfrac{2y}{5}\right)\left(\dfrac{9x^2}{y^2}-\dfrac{6x}{5}+\dfrac{4y^2}{25}\right)\)

Ta có: \(\dfrac{27x^3}{y^3}+\dfrac{8y^3}{125}\)

\(=\left(\dfrac{3x}{y}\right)^3+\left(\dfrac{2y}{5}\right)^3\)

\(=\left(\dfrac{3x}{y}+\dfrac{2y}{5}\right)\cdot\left(\dfrac{9x^2}{y^2}-\dfrac{6xy}{5y}+\dfrac{4y^2}{25}\right)\)

\(=\left(\dfrac{3x}{y}+\dfrac{2y}{5}\right)\left(\dfrac{9x^2}{y^2}-\dfrac{6x}{5}+\dfrac{4y^2}{25}\right)\)

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

8x^3+1=(2x+1)(4x^2-2x+1)

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

125-x^3=(5-x)(25+5x+x^2)

x^3+8y^3=x^3+(2y)^3

=(x+2y)(x^2-2xy+4y^2)

64y^3-125x^3

=(4y)^3-(5x)^3

=(4y-5x)(16y^2+20xy+25x^2)

\(27x^3-\dfrac{1}{8}=\left(3x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(3x-\dfrac{1}{2}\right)\left(9x^2+\dfrac{3}{2}x+\dfrac{1}{4}\right)\)

\(a^6-b^3=\left(a^2\right)^3-b^3\)

\(=\left(a^2-b\right)\cdot\left(a^4+a^2b+b^2\right)\)

\(27x^3\)\(+\)\(8y^3\)\(=\)\(3^3\times x^3\)\(+\)\(2^3\times y^3\)\(=\)\((3x)^3\)\(+\)\((2y)^3\)\(=\)\((3x+2y)\)\(\times\)\((3^2\times x^2-3x\times2y+2^2\times y^2)\)

Khai triển các tích sau

a) (3x+5)^2                         e) 27y^3-8

b) (x^2-4y)^2

c) (8y+1) (8y-1)

d) (2x^3+1)^3

f) 125+27y^3                                  

a) ( 3x + 5 )² = 9x²+30x+25

b) ( x²- 4y )² = x⁴ - 8x²y + 16y²

c) ( 8y+1 )( 8y-1 ) = 64y² - 1

d) ( 2x³+1 ) = 8x⁹+6x⁶+6x³+1

e) 27y³ - 8 = ( 3y )³ - 2³ = ( 3y -2 )( 9y²+6y+4 )

f)125 + 27y³ = 5³ + ( 3y )³ = ( 5+3y )( 25+30y+9y² )

Hk tốt

Bạn sai đề rồi :

 a ) \(8y^3-125\)

\(=\left(2y\right)^3-5^3\)

\(=\left(2y-5\right)\left(4y^2+2y.5+5^2\right)\)

\(=\left(2y-5\right)\left(4y^2+10y+25\right)\)

b) Ta thấy \(8z^3=\left(2z\right)^3\)còn \(27=3^3\)

Trở thành : \(\left(2z\right)^3+3^3\)

Rồi bạn bạn tự làm nha

Thanks

1. x³ + 8 = (x + 2 )(x² - x + 1)

2. 27 - 8y³ = ( 3 - 2y ) ( 9 + 6y + 4y² )

3. y⁶ + 1 = (y²)³ + 1 = ( y² + 1) ( y⁴ - y² +1 )

4.64x³ - \(\dfrac{1}{8}\)y³ = ( 4x - \(\dfrac{1}{2}\)y ) ( 16x² + 2xy + \(\dfrac{1}{4}\)y²)

5. 125x⁶ - 27y⁹ = (5x²)³ - (3y³)³

= ( 5x² - 3y³)(25x⁴ +15x²y³ + 9y⁶)

A= x³+125 = x³+5³ = (x+5)(x²-5x+25)

B= 8y³ -1= (2y)³ - 1³ = (2y-1)(4y²+2y+1)

C= 64x³+27 = (4x)³ +3³ = (4x+3)(16x²-12x+9)

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

b) \(a^6-b^3=\left(a^2\right)^3-b^3=\left(a^2-b\right)\left(b^2+a^2b+a^4\right)\)

c) \(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)

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

a) x³+8y³x3+8y3

=(x+2y)(x²−2xy+4y²)

b) a⁶−b³

=(a²)³-b³

=(a²-b) (a⁴+a²b+b²)

c) 8y³−125

=(2y−5)(4y²+10y+25)

d) 8x³+27

=(2x+3)(4x²−6x+9)

hok tốt!!!

sorry bn có mấy câu có mũ mk ko vết

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

\(8y^3-125\\ =\left(2y\right)^3-5^3\\ =\left(2y-5\right)\left(4y^2+10y+25\right)\)

\(a^6-b^3\\ =\left(a^2\right)^3-b^3\\ =\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)

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

\(x^{32}-1\\ =\left(x^{16}\right)^2-1^2\\ =\left(x^{16}-1\right)\left(x^{16}+1\right)\\ =\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

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

\(x^2-20x+100\\ =\left(x-10\right)^2\)

\(y^4-14y^2+49\\ =\left(y^2-7\right)^2\)

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

b ) \(a^6-b^3=\left(a^2\right)^3-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)

c ) \(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)

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

a) x³ + 8y³ = x³ + (2y)³ = (x+2y)(x²+2xy+4y²)

b) a⁶ - b³ = (a²)³ - b³ = (a²-b)(a⁴ + a²b + b²)

c) 8y³ - 125 = (2y)³ - 5³ = (2y - 5)(4y² + 10y + 25)

d) 8x³ + 27 = (2z)³ + 3³ = (2z + 3)(4z² - 6x + 9)

a) \(x^3+8y^3\)

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

\(=\left[x+2y\right]\left[x^2-x.2y+\left(2y\right)^2\right]\)

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