viết thành một tích
a) a3 + b3
b) x6 + 1
c) 125x3 - 1/8
d) 0,001x3 - 27y3
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Những câu hỏi liên quan
3 tháng 10 2021
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: \(64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
26 tháng 7 2023
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)
MP
26 tháng 7 2023
\(a.x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
\(b.27x^3+1=\left(3x+1\right)\left(9x-3x+1\right)\)
\(c.x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
\(d.64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
T
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, \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)
$Toru$
25 tháng 8 2021
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
21 tháng 8 2021
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: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
CB
1
H9
HT.Phong (9A5)
CTVHS
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
20 tháng 7 2023
`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)`.
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
20 tháng 7 2023
\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
TC
31 tháng 7 2021
1. ( 3x + 2)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)
31 tháng 7 2021
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) \(y^4-14y^2+49=\left(y^2-7\right)^2\)
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)
k cho mính nha