C

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 ...

khong biet

( 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

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) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

a)\(9x^2-6x-3=0\)

\(\Leftrightarrow\)\(3x^2-2x-1=0\)

\(\Leftrightarrow\)\(3x^2-3x+x-1=0\)

\(\Leftrightarrow\)\((3x-1)(x-1)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=1\\ x=-\dfrac{1}{3} \end{array} \right.\)

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) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)

\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)

a) \(9x^2-6x-3=0\\ \Rightarrow\left(9x^2-9x\right)+\left(3x-3\right)=0\\ \Rightarrow9x\left(x-1\right)+3\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(9x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\9x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(x^3+9x^2+27x+19=0\\ \Rightarrow\left(x^3+x^2\right)+\left(8x^2+8x\right)+\left(19x+19\right)=0\\ \Rightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x^2+8x+19=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\\left(x^2+8x+16\right)+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\\left(x+4\right)^2+3=0\left(vôlí\right)\end{matrix}\right.\)

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: \(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{9x^2+6x+4}=3x-2\)

a: \(=\dfrac{2x^3+10x^2-31x^2-155x+222x+1110-1170}{x+5}\)

\(=2x^2-31x+222+\dfrac{-1170}{x+5}\)

c: \(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{9x^2+6x+4}=3x-2\)