ĐKXĐ: `x > -1` và `x ne 0`
`x^2/(x+ 1 - sqrt(x+1))^2 = (x^2 + 3x +18)/(x + 1)^2`
`x/(x + 1 -sqrt(x+ 1)) = (x(x + 1 + sqrt(x +1)))/((x + 1)^2 - (x+ 1))`
`= (x(x +1 +sqrt(x + 1)))/(x(x +1)) = (x+ 1 + sqrt(x +1))/(x +1)` (vì `x ne 0`)
`->((x + 1 +sqrt(x + 1))/(x+ 1))^2 = (x^2 + 3x + 18)/(x +1)^2`
`iff (x + 1 + sqrt(x+ 1))^2 = x^2 + 3x +18`
`iff (x + 1)^2 + 2(x+ 1)sqrt(x + 1) + (x +1) = x^2 +3x + 18`
`iff x^2 + 3x +2 + 2(x + 1)sqrt(x +1) = x^2 + 3x +18`
`iff 2(x+ 1)sqrt(x + 1) =16`
`iff (x + 1)sqrt(x+ 1) - 8 = 0`
`iff ((x +1)^3 - 64)/((x+ 1)sqrt( x + 1) + 8) = 0`
Do `(x + 1)sqrt(x +1) + 8 > 0 forall x > -1` nên:
`(x+ 1)^3 - 64 = 0`
`iff (x + 1)^3 = 64`
`iff x +1 = 4 iff x = 3` (t/m)
Vậy `S={3}`