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\(a,x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
\(b,27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
\(c,y^6+1=\left(y^2\right)^3+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
\(d,64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
\(e,125x^6-27y^9=\left(5x^2\right)^3-\left(3y^3\right)^3=\left(5x^2-3y^3\right)\left(25x^4+15x^2y^3+9y^9\right)\)
\(g,16x^2\left(4x-y\right)-8y^2\left(x+y\right)+xy\left(16+8y\right)\)
\(=8\left[2x^2\left(4x-y\right)-y^2\left(x+y\right)\right]+8xy\left(2+y\right)\)
\(=8\left(8x^3-2x^2y-xy^2-y^3+2xy+xy^2\right)\)
\(f,-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left[\left(\dfrac{x^2}{5}\right)^3+\dfrac{y^3}{4^3}\right]=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
D) 64x^3-1/8y^3
= (4x)^3 + (1/2y)^3
= ( 4x + 1/2y ) [ (4x)^2 - 4x.1/2y + (1/2y)^2 ]
E) 125x^6-27y^9
( câu này mik chưa rõ nên vx chưa tek giải cho bn )
HOk tốt nhé
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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) \(=\left(x+3\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(8-12x+6x^2-x^3\)
\(=-x^3+8+6x^2-12x\)
\(=-\left(x^3-2^3\right)+6x\left(x-2\right)\)
\(=-\left(x-2\right)\left(x^2+2x+4\right)+6x\left(x-2\right)\)
\(=\left(x-2\right)\left(-x^2-2x-4+6x\right)\)
\(=\left(x-2\right)\left(-x^2+4x-4\right)\)
\(=-\left(x-2\right)\left(x-2\right)^2\)
\(=-\left(x-2\right)^3\)
b) \(48x+64+x^3+12x^2\)
\(=x^3+3.4.x^2+3.x.4^2+4^3\)
\(=\left(x+4\right)^3\)
c) \(-9y^2+y-\dfrac{1}{27}+27y^3\)
\(=27y^3-9y^2+y-\dfrac{1}{27}\)
\(=\left(3y\right)^3-3.\left(3y\right)^2.\dfrac{1}{3}+3.3y.\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3\)
\(=\left(3y-\dfrac{1}{3}\right)^3\)
d) \(8x^3+150x-125-60x^2\)
\(=8x^3-60x^2+150x-125\)
\(=\left(2x\right)^3-3.\left(2x\right)^2.5+3.2x.5^2-5^3\)
\(=\left(2x-5\right)^3\)
a, \(8-12x+6x^2-x^3=-\left(x^3-6x^2+12x-8\right)\)
\(=-\left(x^3-2x^2-4x^2+8x+4x-8\right)\)
\(=-\left(x-2\right)^3\)
b, \(48x+64+x^3+12x^2=x^3+4x^2+8x^2+32x+16x+24\)
\(=\left(x+4\right)^3\)
c, \(-9y^2+y-\dfrac{1}{7}+27y^3\)
(sai đề)
d, \(8x^3+150x-125-60x^2=8x^3-20x^2-40x^2+100x+50x-125\)
\(=4x^2\left(2x-5\right)-20x\left(2x-5\right)+25\left(2x-5\right)\)
\(=\left(2x-5\right)\left(4x^2-20x+25\right)=\left(2x-5\right)\left(2x-5\right)^2\)
\(=\left(2x-5\right)^3\)
Chúc bạn học tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x+2) \(\left(x^2-2x+4\right)\)
b) (3 - 2y) \(\left(9+6y+4y^2\right)\)
d) (4x - y) \(\left(16x^2+4xy+y^2\right)\)
bạn giải chi tiết hộ mình với nha.mk sắp phải nộp bài r. huhuhuhu
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
b) \(27y^3-9y^2+y-\frac{1}{27}=\left(3y-\frac{1}{3}\right)^3\)
c) \(8x^6+12x^4y+6x^2y+y^3=\left(2x^2+y\right)^3\)
d) \(\left(x+y\right)^3\left(x-y\right)^3=\left(x^2-y^2\right)^3\)
e) \(\left(x^2-y^2\right)^2\left(x+y\right)\left(x-y\right)=\left(x^2-y^2\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 : Viết các đa thức sau dưới dạng lập phương của một tổng hoặc lập phương của một hiệu
a,8x3+12x2y+6xy2+y3
= (2x)3 + 3.(2x)2.y + 3.2x.y2 + y3
= ( 2x + y )3
b,x3+3x2+3x+1
= x3 + 3.x2.1 + 3.x.12 + 13
=(x + 1)3
c,x3−3x2+2x−1
= x3 - 3.x2.1+ 3.x.12 - 13
= (x - 1)3
d,6
= 33 + 3.32.y2 + 3.3.y4 + (y2)3
= ( 3 + y2 ) 3
cho hỏi lập phương của 1 tổng hay 1 hiệu hay tổng hiệu 2 lập phương vậy
bn viết đề vậy mk cx bí thui haizzzzzz
b) \(125x^3+27y^3=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
c) \(27-y^3=3^3-y^3=\left(3-y\right)\left(9+3y+y^2\right)\)