`(\sqrt{x+3}-\sqrt{x-1})(1+\sqrt{x^2+2x-3})=4` $(1)$
$ĐK: x\ge 1$
Đặt `a=\sqrt{x+3}\ (a\ge 2)`
`\qquad b=\sqrt{x-1}\ (b\ge 0)`
`=>a^2-b^2=x+3-x+1=4`
`\qquad ab=\sqrt{(x+3)(x-1)}=\sqrt{x^2+2x-3}`
`(1)<=>(a-b)(1+ab)=a^2-b^2`
`<=>(a-b)(1+ab)-(a-b)(a+b)=0`
`<=>(a-b)(1+ab-a-b)=0`
$⇔\left[\begin{array}{l}a-b=0\ (VN)\\1+ab-a-b=0\ (2)\end{array}\right.$
`(2)<=>a(b-1)-(b-1)=0`
`<=>(a-1)(b-1)=0`
`<=>b-1=0` (vì `a\ge 2=>a-1\ne 0)`
`<=>b=1`
`⇔\sqrt{x-1}=1`
`<=>x-1=1`
`<=>x=2 (TM)`
Vậy `S={2}`
