a) Ta có: \(5x\left(x-127\right)-x+127=0\)
\(\Leftrightarrow5x\left(x-127\right)-\left(x-127\right)=0\)
\(\Leftrightarrow\left(x-127\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-127=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=127\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=127\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{127;\frac{1}{5}\right\}\)
b) Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: x∈{3;-2}
a,5x(x-127)-x+127=0 \(^{_{ }\Leftrightarrow}\)5x(x-127)-(x-127)=0 \(\Leftrightarrow\)(5x-1)(x-127)=0 \(\Leftrightarrow\) \(\left[{}\begin{matrix}5x-1=0\\x-127=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\frac{1}{5}\\x=127\end{matrix}\right.\)
b,\(x^2\)-x-6=0 \(\Leftrightarrow\)\(x^2\)+3x-2x-6=0 \(\Leftrightarrow\)(\(x^2\)+3x)-(2x+6)=0 \(\Leftrightarrow\)x(x+3)-2(x+3)=0 \(\Leftrightarrow\)(x+3)(x-2)=0 \(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
