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11 tháng 7 2016

Ta có

a, a3+b3=(a+b)(a2-ab+b2)

b,x6+1=(x2+1)(x2-x+1)

c,125x3-1/8=(5x)3-1/23=(5x-1/2)(25x2+2,5x+1/4)

d,0,001x3-27y3=(0,1)3-(3y)3=(0,1-3y)(0,01+0,3y+y2)

11 tháng 7 2016

k cho mính nha

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

b: =(3x+1)(9x^2-3x+1)

c: =(x+3)(x^2-3x+9)

d: =(4x-3y)(16x^2+24xy+9y^2)

26 tháng 7 2023

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

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

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

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

27 tháng 10 2023

a, \(8^3yz+12^2yz+6xyz+yz\)

\(=512yz+144yz+6xyz+yz\)

\(=yz\left(512+14+6x+1\right)\)

\(=yz\left(527+6x\right)\)

$---$

b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)

\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)

\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)

\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)

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

$---$

c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)

\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)

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

\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)

$---$

d, \(x^6+x^4+x^2y^2+y^4-y^6\)

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

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

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

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

$Toru$

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

25 tháng 8 2021

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

1: \(4x^2+4x+1=\left(2x+1\right)^2\)

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

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

4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

20 tháng 8 2023

a) \(\widehat{B_1}=\widehat{B_3}=55^o\)

Hai góc đối đỉnh

Mà: \(\widehat{B_3}+\widehat{B_4}=180^o\) (kề bù)

\(\Rightarrow\widehat{B_4}=180^o-55^o=125^o\)

Mà: \(\widehat{B_2}=\widehat{B_4}=125^o\) (đối đỉnh)

b) Ta có: a//b

\(\Rightarrow\widehat{B_3}=\widehat{A_3}=55^o\)

Hai góc đồng vị

Mà: \(\widehat{B_2}=\widehat{A_4}=125^o\)

Hai góc so le trong

Mà: \(\widehat{B_1}=\widehat{A_1}=55^o\)

Đồng vị

Mà: \(\widehat{B_2}=\widehat{A_2}=125^o\)

Hai góc đồng vị

`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.

`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.

`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.

20 tháng 7 2023

\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

31 tháng 7 2021

1. ( 3x + 2)- 4

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

= 3x(3x+4)

2. 4x2 - 25y2

= (2x-5y)(2x+5y)

3. 4x2- 49

=(2x-7)(2x+7)

4. 8z3 + 27

=(2z+3)(4x2-6z+9)

5. \(\dfrac{9}{25}x^4-\dfrac{1}{4}\)

\((\dfrac{3}{5}x^2-\dfrac{1}{2})(\dfrac{3}{5}x^2+\dfrac{1}{2})\)

6. x32  - 1

=(x16-1)(x16+1)

7. 4x2 + 4x + 1

=(2x+1)2

8. x2 - 20x + 100

=(x-10)2

9. y4 -14y2 + 49

=(y2-7)2

10.  125x3 - 64y3

= (5x-4y)(25x2+20xy+16y2)

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

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

3) \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)

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

5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

6) \(x^{32}-1=\left(x^{16}-1\right)\left(x^{16}+1\right)\)

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

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

7) \(4x^2+4x+1=\left(2x+1\right)^2\)

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

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