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Giải:
\(\left(27x^3-8\right):\left(6x+9x^2+4\right)\)
\(=\dfrac{27x^3-8}{6x+9x^2+4}\)
\(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{6x+9x^2+4}\)
\(=3x-2\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
( 27\(x^3\) - 8) : ( 9\(x^2\) + 6x + 4)
= [ \(\left(3x\right)^3\) - 23] : ( 9\(x^2\) + 6x + 4)
= (3x - 2)( 9\(x^2\) + 6x + 4) : ( 9\(x^2\) + 6x + 4)
= 3x - 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
b) \(-x^3-3x^2-3x-1=-\left(x^3+3x^2+3x+1\right)=-\left(x+1\right)^3\)
c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)
Tìm x.
a) 9x^2 – 6x – 3 = 0
b) x^3 + 9x^2 + 27x + 19 = 0
c) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(9x^2-6x-3=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x^3+9x^2+27x+19=0\)
\(\Leftrightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)
\(\Leftrightarrow x=-1\)( do \(x^2+8x+19=\left(x+4\right)^2+3>0\))
c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)
\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=8x^3-y^3\)
b: \(=2x^2-3xy+5y^2\)
c: \(=\dfrac{2x^3+10x^2-31x^2-155x+222x+1110-1170}{x+5}\)
\(=2x^2-31x+222+\dfrac{-1170}{x+5}\)
e: \(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{9x^2+6x+4}=3x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{2x^3+10x^2-31x^2-155x+222x+1110-1170}{x+5}\)
\(=2x^2-31x+222+\dfrac{-1170}{x+5}\)
c: \(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{9x^2+6x+4}=3x-2\)
\(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{9x^2+6x+4}=3x-2\)