\(x\left(2-x\right)+\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow2x-x^2+x^2-16=0\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

Để hàm số \(y=\left(2-m\right).x+m-1\) là hàm số bậc nhất thì \(2-m\ne0\Leftrightarrow m\ne2\)

\(\Rightarrow\) Chọn đáp án \(C\)

\(b,\left|x+\dfrac{1}{2}\right|-\dfrac{3}{4}=\dfrac{5}{8}\)

\(\Rightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{11}{8}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{11}{8}\\x+\dfrac{1}{2}=-\dfrac{11}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{15}{8}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{7}{8};-\dfrac{15}{8}\right\}\)

\(a,\dfrac{2}{3}.\left(x+\dfrac{1}{3}\right)=\dfrac{5}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{5}{2}\)

\(\Rightarrow x=\dfrac{13}{6}\)

Vậy: \(x=\dfrac{13}{6}\)