$x^2-2x-8=4\sqrt{(4-x)(x+2)}_{}$ $ĐKXĐ:_{}$ $-2_{}$ $\leq$ $x_{}$ $\leq4$
$⇔x^2-2x-8=4\sqrt{4x+8-x^2-2x}_{}$
$⇔x^2-2x-8=4\sqrt{2x+8-x^2}_{}$
$⇔4\sqrt{2x+8-x^2}=x^2-2x-8_{}$
$⇔16.(2x+8-x^2)=x^4+4x^2+64-4x^3-16x^2+32x_{}$
$⇔32+128-16x^2=x^4+4x^2+64-4x^3-16x^2+32x_{}$
$⇔128=x^4+4x^2+64-4x^3_{}$
$⇔128-x^4-4x^2-64+4x^3=0_{}$
$⇔64-x^4-4x^2+4x^3=0_{}$
$⇔4.(16-x^2)+x^3.(-x+4)=0_{}$
$⇔4.(4-x)(4+x)+x^3.(-x+4)=0_{}$
$⇔(4-x).[ 4.(4+x)+x^3]=0_{}$
$⇔(4-x)(x^3+4x+16)=0_{}$
$⇔(4-x)(x^3+2x^2-2x^2-4x+8x+16)=0_{}$
$⇔(4-x).[ x^2.(x+2)-2x.(x+2)+8.(x+2)]=0_{}$
$⇔(4-x)(x+2)(x^2-2x+8)=0_{}$
$(vì_{}$ $x^2-2x+8∉R)_{}$ $(Loại)_{}$
$⇔_{}$ \(\left[ \begin{array}{l}4-x=0\\x+2=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=4(TMĐK)\\x=-2(TMĐK)\end{array} \right.\)
Vậy phương trình trên có nghiệm $x_{1}=4;$ $x_{2}=-2$
