`4x-4=9+6\sqrt{6-x}+6-x` `(x ≤ 6)`
`⇔5x-19=6\sqrt{6-x}` `(x ≥ 19/5)`
`⇔(5x-19)^2=(6\sqrt{6-x})^2`
`⇔25x^2-190x+361=36(6-x)`
`⇔25x^2-190x+361=216-36x`
`⇔25x^2-154x+145=0`
`⇔25x^2-125x-29x+145=0`
`⇔25x(x-5)-29(x-5)=0`
`⇔(25x-29)(x-5)=0`
`⇔` \(\left[ \begin{array}{l}25x-29=0\\x-5=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{29}{25}(ktm)\\x=5(tm)\end{array} \right.\)
Vậy `x=5`