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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : x 3 - 6 x 2 y + 12 x y 2 - 8 y 3 = ( x ) 3 - 3 . x 2 . 2 y + 3 . x . ( 2 y ) 2 - ( 2 y ) 3 = ( x - 2 y ) 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : x 3 - 6 x 2 y + 12 x y 2 - 8 y 3 = ( x ) 3 - 3 . x 2 . 2 y + 3 . x . ( 2 y ) 2 - ( 2 y ) 3 = ( x - 2 y ) 3
a,x3+3x2+3x+1
b,x2+6x+9
c,-x3+9x2-27x+27
d,x2+4x+4
k,10x-25-x2
f,(x+y)2-9x2
g,8x3+42x2y+16xy2+6xy+y3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)
b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-\left(x^3-9x^2+27x-27\right)\)
\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)
d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)
f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)
\(=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)