ĐK:$x\geq1$${x^2} - 8\left( {x + 3} \right)\sqrt {x - 1} + 22x - 7 = 0$
$<=>(x^2+22x-7-128)+(128-8(x+3)\sqrt{x-1})=0$
$<=>(x-5)(x+27)+8.\frac{(x+3)^2(x-1)-16^2}{(x+3)\sqrt{x-1}+16}=0$
$<=>(x-5)(x+27)+8\frac{(x-5)(x^2+10x+53)}{(x+3)\sqrt{x-1}+16}=0$
$<=>x=5$ hoặc $(x+27)+8\frac{x^2+10x+53}{(x+3)\sqrt{x-1}+16}=0(1)$
với đk $x\geq1$
$=>(1)>0$
vậy pt có 1 nghiệm: $x=5$