Bài 2:
a: 4x(x-3)+6(3-x)=0
=>4x(x-3)-6(x-3)=0
=>(x-3)(4x-6)=0
=>\(\left[{}\begin{matrix}x-3=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(x^3-x\left(x+1\right)\left(x-1\right)=14\)
=>\(x^3-x\left(x^2-1\right)=14\)
=>\(x^3-x^3+x=14\)
=>x=14
c: \(\left(x^2-x\right)^2+2\left(x^2-x\right)=8\)
=>\(\left(x^2-x\right)^2+2\left(x^2-x\right)-8=0\)
=>\(\left(x^2-x\right)^2+4\left(x^2-x\right)-2\left(x^2-x\right)-8=0\)
=>\(\left(x^2-x\right)\left(x^2-x+4\right)-2\left(x^2-x+4\right)=0\)
=>\(\left(x^2-x+4\right)\left(x^2-x-2\right)=0\)
=>\(\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\right)\left(x-2\right)\left(x+1\right)=0\)
=>\(\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
