\(a,3x\left(x^2-4\right)=0\\ 3x\left(x-2\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
\(b,x\left(x-2\right)-x+2=0\\ x\left(x-2\right)-\left(x-2\right)=0\\ \left(x-1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(c,x^2\left(x^2-1\right)-x^2-1=0\\ x^4-2x^2+1-2=0\\ \left(x^2-1\right)^2=2\\ \left[{}\begin{matrix}x^2-1=\sqrt{2}\\x^2-1=-\sqrt{2}\end{matrix}\right.\left(chịu\right)\)
\(d,5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\\ 5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\\ 5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\\ 30x-55=5\\ 30x=60\\ x=60:30\\ x=2\)