a/ \(x^4+2x^3-4x^2-5x-6=0\)
\(\Leftrightarrow x^4+x^3+x^2-2x^3-2x^2-2x+3x^3+3x^2+3x-6x^2-6x-6=0\)
\(\Leftrightarrow x^2\left(x^2+x+1\right)-2x\left(x^2+x+1\right)+3x\left(x^2+x+1\right)-6\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-2x+3x-6\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x-2\right)\left(x+3\right)=0\)
Vì \(x^2+x+1>0\forall x\Rightarrow\) vô nghiệm
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy pt có 2 nghiệm-
b/ \(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4+3x^3-2x^3-18x^3-9x^2+6x^2+3x-12x^3-6x^2+4x^2+2x+36x^2+18x-12-6=0\)
\(\Leftrightarrow3x^3\left(2x+1\right)-x^2\left(2x+1\right)-9x^2\left(2x+1\right)+3x\left(2x+1\right)-6x^2\left(2x+1\right)+2x\left(2x+1\right)+18x\left(2x+1\right)-6\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x^3-x^2-9x^2+3x-6x^2+2x+18x-6\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left[x^2\left(3x-1\right)-3x\left(3x-1\right)-2x\left(3x-1\right)+6\left(3x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-1\right)\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\3x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\\x=2\\x=3\end{matrix}\right.\)
