\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)
\(\Leftrightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow-3x+2=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy .............
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(x^3-3^3-3x+17=x^3-12\)
\(-3x=-12+27-17=-2\)
=>\(x=\dfrac{2}{3}\)
Vậy...
\(( x- 3) ( x^2 + 3x +9) - ( 3x - 17) = x^3 - 12\)\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow-3x+2=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
