a) `6``.``x``=``-``54`
`x``=``-``54``:``6`
`x``=``-``9`
b) `-``8``.``x``=``-``72`
`x``=``-``72``:``(``-``8``)`
`x``=``9`
c) `x²``-``5x``=``0`
`x``.``(``5``+``x``)``=``0`
⇒ $\left[ \begin{array}{l}x=0\\5+x=0\end{array} \right.$⇒$\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.$
Vậy `x``∈``(``0``;``-``5``)`
d) `5``.``(``x`` -``2``)``=``0`
`⇒` x`-``2``=`0
`x``=``0``+``2`
`x``=``2`
e) `5``.``(``-``4``)``.``x``=``-``100`
`-``20``.``x``=``-``100`
`x``=``-``100``:``(``-``20``)`
`x``=``5`
f) `-``5``.``x``=``-``6``.``(``-``10``)`
`-``5``.``x``=``60`
`x``=``60``:``(``-``5)`
`x``=``-``12`
