a) \(-x^3+9x^2-27x+27=-\left(x^3-3.3.x^2+3.3^2.x-3^3\right)=-\left(x-3\right)^3\)

b)\(x^4-2x^3-x^2+2x+1=x^4+\left(-x\right)^2+\left(-1\right)^2+2x^2\left(-x\right)+2.\left(-x\right).\left(-1\right)+2x^2.\left(-1\right)\)

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

c)\(8x^3+27y^3+36x^2y+54xy^2=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)

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

Bài 1: 

a: \(C=\left(x-3\right)\left(x+3\right)-\left(x+5\right)\left(x-1\right)\)

\(=x^2-9-\left(x^2+4x-5\right)\)

\(=x^2-9-x^2-4x+5=-4x-4\)

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

\(=\left(3x-2+x+1\right)^2=\left(4x-1\right)^2=16x^2-8x+1\)

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

2, đề sai 

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

4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)

5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)

tương tự ... 

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

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

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

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

Sửa lại câu d) là `25y^2`

`a)x^3-1`

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

`b)8x^3-y^3`

`=(2x)^3-y^3`

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

`c)x^2-8x+16`

`=x^2-2.x.4+4^2`

`=(x-4)^2`

`d)25y^2-1`

`=(5y)^2-1`

`=(5y-1)(5y+1(`

`e)27-8y^3`

`=3^3-(2y)^3`

`=(3-2y)(9+6y+4y^2)`

`f)2x^2-8x+8`

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

`=2(x-2)^2`

a) x³ - 1 = x³ - 1³

              = (x - 1)(x² - x + 1)

b) 8x³ - y³ = (2x)³ - y³

                 = (2x - y)(4x² + 2xy + y²)

c) x² - 8x + 16 = x² - 2.4x + 4²

                        = (x - 4)²

d) đề có bị sai không , nên mình sưa lại đề nhé :

       25y² - 1 = (5y)² - 1²

                     = (5y - 1)(5y + 1)

e) 27 - 8y³ = 3³ - (2y)³

                  = (3 - 2y)(9 + 6y + 4y²)

f) 2x² - 8x + 8 = 2(x² - 4x + 4)

                       = 2(x - 2)²

 Chúc bạn học tốt

MỌI NGƯỜI TRẢ LỜI GIÚP MÌNH VỚI MÌNH CẦN GẤP LẮP

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

2.\(8x^6-27y^3=\left(2x\right)^3-\left(3y\right)^3=\left(2x-3y\right)\)

\(=\left(2x-3y\right)\left(8x^6+6xy+27y^3\right)\)

3.\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)4.câu cuối là \(8b^3\)bạn nhé !!

tik mik nhé

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

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

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

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

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

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

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

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

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

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

1: Ta có: \(x^{10}-4x^8+4x^6\)

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

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

2: Ta có: \(m^3+27\)

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

3: Ta có: \(x^3+8\)

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

4: Ta có: \(\frac{1}{27}+a^3\)

\(=\left(\frac{1}{3}+a\right)\left(\frac{1}{9}-\frac{a}{3}+a^2\right)\)

5: Ta có: \(8x^3+27y^3\)

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

6: Ta có: \(\frac{1}{8}x^3+8y^3\)

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

7: Ta có: \(8x^6-27y^3\)

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

8: Ta có: \(\frac{1}{8}x^3-8\)

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

9: Ta có: \(\frac{1}{64}x^6-125y^3\)

\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)

10: Ta có: \(\left(a+b\right)^3-c^3\)

\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)\cdot c+c^2\right]\)

\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)

11: Ta có: \(x^3-\left(y-1\right)^3\)

\(=\left[x-\left(y-1\right)\right]\cdot\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]\)

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

12: Ta có: \(x^6+1\)

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

1) \(x^{10}-4x^8+4x^6\)

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

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

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

4) \(\frac{1}{27}+a^3=\left(\frac{1}{3}\right)^3+a^3=\left(\frac{1}{3}+a\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}a+a^2\right]\)

5) \(8x^3+27y^3=\left(2x\right)^3+\left(3y\right)^3=\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

6) \(\frac{1}{8}x^3+8y^3=\left(\frac{1}{2}x\right)^3+\left(2y\right)^3=\left(\frac{1}{2}x+2y\right)\left[\left(\frac{1}{2}x\right)^2-\frac{1}{2}x.2y+\left(2y\right)^2\right]=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)

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

10) \(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left[\left(a^2+2ab+b^2\right)+ac+bc+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)11) \(x^3-\left(y-1\right)^3=\left(x-y+1\right)\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]=\left(x-y+1\right)\left[x^2+xy-x+\left(y^2-2y+1\right)\right]=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)

P/s: Đăng ít thôi chớ bạn!