`\qquad x^2-2(x+1)sqrt{x^2-1}-3x^2+6x-1=0` ĐK:`[(x>=1),(x<=-1):}`
`<=> (x^2+2x+1)-2(x+1)sqrt{x^2-1}+x^2-1=4x^2-4x+1`
`<=> (x+1-sqrt{x^2-1})^2=(2x-1)^2`
`<=> (2x-1)^2-(x+1-sqrt{x^2-1})^2=0`
`<=> (2x-1-x-1+sqrt{x^2-1})(2x-1+x+1-sqrt{x^2-1})=0`
`<=> (x-2+sqrt{x^2-1})(3x-sqrt{x^2-1})=0`
`<=> [(sqrt{x^2-1}=2-x (1)),(sqrt{x^2-1}=3x (2)):}`
Giải pt(1): `sqrt{x^2-1}=2-x` ĐK: `1<=x<=2`
`<=> x^2-1=4-4x+x^2`
`<=> 4x=5`
`<=> x=5/4 (\text{tm})`
Giải pt (2): `sqrt{x^2-1}=3x` ĐK:`x>=1`
`<=> x^2-1=9x^2`
`<=> 8x^2=-1 (\text{vô lý})`
Vậy `S={5/4}`
