a) \(x^5-27+x^3-27x^2\) = 0
\(\Leftrightarrow x^3\left(x^2+1\right)-27\left(x^2+1\right)\)= 0
\(\Leftrightarrow\left(x^2+1\right)\left(x^3-27\right)=0\)
\(\Leftrightarrow x^3-27=0\) (Vì \(x^2+1>0\))
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+2\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{27}{4}\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}\right]=0\)
\(\Leftrightarrow x-3=0\) (Vì \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}>0\))
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của phương trình là S = {3}
b)\(x^3-9x^2+19x-11=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(8x^2-8x\right)+\left(11x-11\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-8x+11\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-\left(4+\sqrt{5}\right)x-\left(4-\sqrt{5}\right)x+11\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left\{x\left[x-\left(4+\sqrt{5}\right)\right]-\left(4-\sqrt{5}\right)\left[x-\left(4+\sqrt{5}\right)\right]\right\}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4-\sqrt{5}\right)\left(x-4+\sqrt{5}\right)=0\)
\(\Leftrightarrow x-1=0\) hoặc \(x-4-\sqrt{5}=0\) hoặc \(x-4+\sqrt{5}=0\)
\(\Leftrightarrow x=1\) hoặc \(x=4+\sqrt{5}\) hoặc \(x=4-\sqrt{5}\)
Vậy phương trình có tập nghiệm là \(S=\left\{1;4+\sqrt{5};4-\sqrt{5}\right\}\)