\(x^2-49+9y^2-6xy\)

\(=\left(x^2-6xy+9y^2\right)-49\)

\(=\left(x-3y\right)^2-49\)

\(=\left(x-3y-7\right)\left(x-3y+7\right)\)

\(16x^3-25x=0\)

\(\Leftrightarrow x\left(16x^2-25\right)=0\)

\(\Leftrightarrow x\left(4x-5\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x-5=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\pm\dfrac{5}{4}\right\}\)

\(6,\)

\(a,x+2017=-1\)

\(\Rightarrow x=-2018\)

Vậy: \(x=-2018\)

\(b,y-\left(-100\right)=1\)

\(\Rightarrow y+100=1\)

\(\Rightarrow y=-99\)

Vậy: \(y=-99\)

\(x^2-10x+24=0\)

\(\Leftrightarrow x^2-10x+25-1=0\)

\(\Leftrightarrow\left(x-5\right)^2-1=0\)

\(\Leftrightarrow\left(x-5-1\right)\left(x-5+1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)

Vậy: \(S\in\left\{6;4\right\}\)

\(8x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(8x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\8x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{8}\end{matrix}\right.\)

Vậy: \(S=\left\{3;\dfrac{1}{8}\right\}\)