b) `(x+2)(x^2-2x+4)-x(x^2+2)=15`
⇔`x^3+8-x^3-2x=15`
⇔`-2x=15-8`
⇔`-2x=7`
⇔`x=-7/2`
Vậy `S={-7/2}`
d) `6x^2-(2x+5)(3x-2)=7`
⇔`6x^2-6x^2+4x-15x+10=7`
⇔`-11x=7-10`
⇔`-11x=-3`
⇔`x=3/(11)`
Vậy `S={3/(11)}`
e) `(5-3x)^2-36=0`
⇔`(5-3x-6)(5-3x+6)=0`
⇔`(-3x-1)(-3x+11)=0`
⇔\(\left[ \begin{array}{l}-3x-1=0\\-3x+11=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{1}{3}\\x=\dfrac{11}{3}\end{array} \right.\)
Vạy `S={-1/3,(11)/3}`
