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\(8x^6-27y^3\)
\(=\left(2x^2\right)^3-\left(3y\right)^3\)
\(=\left(2x^2-3y\right)\left(4x^2+6x^2y+9y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
Câu 2 đề ko sai nha bạn.
2) x2 - (\(\sqrt{y^3}\))2 ( y>0)
= ( x -\(\sqrt{y^3}\)) ( x +\(\sqrt{y^3}\))
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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)
a) (1 - 2x) (2x + 1) = 1 - 4x2 __ hằng đẳng thức số 3 (A + B) (A - B) = A2 - B2 (ở đây A = 1 , B = 2x)
câu b) có sai đề ko bn
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a)(x+3y)(x2-3xy+9y2)=x3+27y3
b)(2x-3y)(4x2+6xy+9y2)=8x3-27y3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy \(8x^3+27y^3\)
\(=\left(2x\right)^3+\left(3y\right)^3\)
\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
\(=4x^2-6xy+9y^2\)
Thế thì \(A=6x^2-6xy+18y^2+5\)
Rồi lại thay \(x=\dfrac{1-3y}{2}\) vào A thôi.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)
`a)8x^3+27y^3`
`=(2x)^3+(3y)^3`
`=(2x+3y)(4x^2-6xy+9y^2)`
8x^3+27y^3
=(2x)^3+(3y)^3
=(2x+3y)((2x)^2−6xy+(3y)^2)