Đáp án+Giải thích các bước giải:
`a)`
`6x(x-2)-x+2=0`
`⇔6x(x-2)-(x-2)=0`
`⇔(x-2)(6x-1)=0`
\(⇔\left[ \begin{array}{l}x-2=0\\6x-1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\6x=1\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\x=\dfrac{1}{6}\end{array} \right.\)
Vậy `x=2` hoặc `x=1/6`
`b)`
`x^2-16=5x(x+4)`
`⇔x^2-16-5x(x+4)=0`
`⇔(x-4)(x+4)-5x(x+4)=0`
`⇔(x+4)(x-4-5x)=0`
`⇔(x+4)(-4x-4)=0`
\(⇔\left[ \begin{array}{l}x+4=0\\-4x-4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-4\\-4x=4\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-4\\x=-1\end{array} \right.\)
Vậy `x=-4` hoặc `x=-1`
