`x+4\sqrt(x+3)+2\sqrt(3-2x)=11`
`⇔(x-1)+4(\sqrt(x+3)-2)+2(\sqrt(3-2x)-1)=0`
`⇔(x-1)+4.(x-1)/(\sqrt(x+3)+2)-2.(x-1)/(\sqrt(3-2x)+1)=0`
`⇔(x-1)(1+4/(\sqrt(x+3)+2)-2/(\sqrt(3-2x)+1))=0`
- Xét `1+4/(\sqrt(x+3)+2)-2/(\sqrt(3-2x)+1)=0`
`⇔(\sqrt(3-2x)+1)(\sqrt(x+3)+2)+\sqrt(3-2x)-\sqrt(x+3)-1=0`
`⇔\sqrt[(3-2x)(x+3)]+3\sqrt(x+3)+1=0` (vô lí)
Nên `⇔x-1=0`
`⇔x=1`
Vậy `x=1`