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\(\left(\dfrac{1}{3}y+3\right)^3=\dfrac{1}{27}y^3+y^2+9y+27\)
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\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1}{\left(3y+3\right)^3}=\dfrac{1}{27y^3+81y^2+81y+27}\)
\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1^3}{\left(3y+3\right)^3}=\dfrac{1}{27\left(y^3+3y^2+3y+1\right)}\)
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a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3\)
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\(-4a^2x\cdot\left(-2bxy\right)^2\cdot\left(-\dfrac{1}{4}x^2y^3\right)\)
\(=-4a^2x\cdot4b^2x^2y^2\cdot\left(-\dfrac{1}{4}x^2y^3\right)\)
\(=\left(-4a^2\cdot4b^2\cdot-\dfrac{1}{4}\right)\left(x\cdot x^2\cdot x^2\right)\left(y^2\cdot y^3\right)\)
\(=4a^2b^2x^5y^5\)
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\(\left(x+y\right)^3=x^3+3x^2y+3xy^2-y^3\)
\(\left(x-y\right)^3=x^3-3x^2y+3xy^2-y^3\)
\(\left(2y-3\right)^3=8y^3-36y^2+54y-27\)
a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(x + y + z) . (x + y) + (x + y)2
=(y+x)z+2y2+4xy+2x2
=(y+x)(z+2y+2x)