d) \(x+\dfrac{2}{4}=\dfrac{3}{6}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x=0\)
Vậy: \(x=0\)
e) \(-\dfrac{1}{5}=\dfrac{4}{10}x\)
\(\Rightarrow-\dfrac{1}{5}=\dfrac{2x}{5}\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
f) Đề là \(\dfrac{x+2}{3}=\dfrac{3}{x+2}\)?
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-5\right\}\)
\(d,x=\dfrac{3}{6}-\dfrac{2}{4}=0\\ e,x=\dfrac{-1}{5}:\dfrac{4}{10}=\dfrac{-1}{2}\\ f,\left(x+2\right)\left(x+2\right)=3.3\\ \left(x+2\right)^2=9\\ x+2=3;x+2=-3\\ x=1;x=-5\)
