\(A\left(x\right)=\left(x^2-x^2\right)-3x+1=-3x+1\)
cho A(x) = 0
\(=>-3x+1=0=>-3x=-1=>x=\dfrac{1}{3}\)
cho B(x) = 0
\(=>6-\dfrac{1}{3}x=0=>\dfrac{1}{3}x=6=>x=6\times3=18\)
Cho C(x)=0
\(=>x^2+2x=0=>x\left(x+2\right)=0=>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
cho D(x) =0
\(=>4x^2-1=0=>4x^2=1=>x^2=\dfrac{1}{4}=>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
cho E(x)=0
\(=>2x^2+3x=0=>x\left(2x+3\right)=0=>\left[{}\begin{matrix}x=0\\2x=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
