c.\(2x^4-5x^2-3=0\)\(\Leftrightarrow2x^4-6x^2+x^2-3=0\)\(\Leftrightarrow2x^2\left(x^2-3\right)+x^2-3=0\)\(\Leftrightarrow\left(2x^2+1\right)\left(x^2-3\right)=0\)\(\Leftrightarrow x^2-3=0\) (do \(2x^2+1>0;\forall x\))\(\Leftrightarrow x=\pm\sqrt{3}\)b.\(3x^3-5x^2+4x-12=0\)\(\Leftrightarrow\left(3x^3-6x^2\right)+\left(x^2-2x\right)+\left(6x-12\right)=0\)\(\Leftrightarrow3x^2\left(x-2\right)+x\left(x-2\right)+6\left(x-2\right)=0\)\(\Leftrightarrow\left(x-2\right)\left(3x^2+x+6\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x^2+x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\3\left(x+\dfrac{1}{6}\right)^2+\dfrac{71}{12}=0\left(vô-nghiệm\right)\end{matrix}\right.\)Câu trả lời: c.\(2x^4-5x^2-3=0\)\(\Leftrightarrow2x^4-6x^2+x^2-3=0\)\(\Leftrightarrow2x^2\left(x^2-3\right)+x^2-3=0\)\(\Leftrightarrow\left(2x^2+1\right)\left(x^2-3\right)=0\)\(\Leftrightarrow x^2-3=0\) (do \(2x^2+1>0;\forall x\))\(\Leftrightarrow x=\pm\sqrt{3}\)b.\(3x^3-5x^2+4x-12=0\)\(\Leftrightarrow\left(3x^3-6x^2\right)+\left(x^2-2x\right)+\left(6x-12\right)=0\)\(\Leftrightarrow3x^2\left(x-2\right)+x\left(x-2\right)+6\left(x-2\right)=0\)\(\Leftrightarrow\left(x-2\right)\left(3x^2+x+6\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x^2+x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\3\left(x+\dfrac{1}{6}\right)^2+\dfrac{71}{12}=0\left(vô-nghiệm\right)\end{matrix}\right.\)