\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow-10-3x=-12\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy...
\((x-3)(x^2+3x+9)-(3x-17)=x^3-12 \)
\(pt\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow x^3-3x-10-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)\(\Leftrightarrow x=\dfrac{2}{3}\)