ĐKXĐ: $x\geq1$
$x-\sqrt{x-1}=3$
`<=>`$\sqrt{x-1}=x-3$
`<=>`$\left \{ {{x-3\geq0} \atop {x-1=(x-3)^{2}}} \right.$
`<=>`$\left \{ {{x\geq3} \atop {x-1=x^{2}-6x+9}} \right.$
`<=>`$\left \{ {{x\geq3} \atop {x^{2}-7x+10=0}} \right.$
`<=>`$\left \{ {{x\geq3} \atop {(x-5)(x-2)=0(1)}} \right.$
Giải $(1)$:
$(x-5)(x-2)=0$
`<=>`\(\left[ \begin{array}{l}x-5=0\\x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=5 (t/m)\\x=2 (loại)\end{array} \right.\)
Vậy pt có nghiệm $x=5$