a) Để \((\dfrac{1}{2}\times x-3)\times(-\dfrac{1}{3}+x)=0\) thì

\(\left[{}\begin{matrix}\dfrac{1}{2}\times x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}\times x=3\\x=\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\div\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy x=6 hoặc x=1/3

d) Ta có: \(9\times x^2=1\)

=> \(x^2=\dfrac{1}{9}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)

Vậy x=1/3 hoặc x=-1/3

e)Ta có: \(\left(x-5\right)^2=4\)

=> \(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy x=7 hoặc x=3

vì đó là bệnh viện thú y