a)\(4x^4+12x^3+5x^2-6x-15=0\) ⇔(x-1)(2x+5)(2\(x^2\)+3x+3)=0\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+5=0\\2x^2+3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-5}{2}\\x=\varnothing\end{matrix}\right.\)

Vậy phương trình đã cho có tập nghiệm S=\(\left\{1;\dfrac{-5}{2}\right\}\)

b)(x+1)(x+2)(x+4)(x+5)=40 ⇔(x+1)(x+5)(x+2)(x+4)-40=0 ⇔(\(x^2\)+6x+5)(\(x^2\)+6x+8)-40=0 Đặt \(x^2+6x+5=a \) ta có: a(a+3)-40=0⇔\(a^2\)+3a-40=0⇔\((a^2-5a)+(8a-40)=0\) ⇔a(a-5)+8(a-5)=0⇔(a-5)(a+8)=0 \(\Leftrightarrow\left[{}\begin{matrix}a-5=0\\a+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+6x+5-5=0\\x^2+6x+5+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+6\right)=0\\x^2+6x+9+2=0\end{matrix}\right.\) \(\circledast x\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\) \(\circledast x^2+6x+9+2=0\Leftrightarrow\left(x^2+6x+9\right)+2=0\Leftrightarrow\left(x+3\right)^2+2=0\Leftrightarrow\left(x+3\right)^2=-2\left(lo\text{ại}\right)\)

Vậy phương trình đã cho có tập nghiệm S=\(\left\{0;-6\right\}\)