a,\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(x^3+27+x^2-6x-27=0\)

\(x^3+x^2-6x=0\)

\(x^3-2x^2+3x^2-6x=0\)

\(\left(x^2+3\right)\left(x-2\right)=0\)

Vì \(x^2+3>0\) nên \(x-2=0=>x=2\)

Vậy...

b,\(x^3-11x^2+30x=0\)

\(x^3-6x^2-5x^2+30x=0\)

\(\left(x-6\right)\left(x^2-5\right)=0\)

=>\(x-6=0\) hoặc \(x^2-5=0\)

=>\(x=6\)hoặc \(x^2=5\)

=>\(x=6\) hoặc \(x=-\sqrt{5}\) hoặc \(x=\sqrt{5}\)

Giải các phương trình sau:

a,\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

b,\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

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

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)\(\Leftrightarrow x=\dfrac{7}{4}\)

Vậy \(S=\left\{\dfrac{7}{4}\right\}\)

b)\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\Leftrightarrow x=\dfrac{9}{8}\)

Vậy \(S=\left\{\dfrac{9}{8}\right\}\)