`x-3\sqrt{x}-5=0` `ĐK: x >= 0`
Đặt `\sqrt{x}=t (t >= 0)` khi đó ptr có:
`t^2-3t-5=0`
Ptr có:`\Delta=(-3)^2-4.(-5)=29 > 0`
`=>` Ptr có `2` nghiệm pb
`t_1=[-b+\sqrt{\Delta}]/[2a]=[3+\sqrt{29}]/2` (t/m)
`t_2=[-b-\sqrt{\Delta}]/[2a]=[3-\sqrt{29}]/2` (ko t/m)
`@t=[3+\sqrt{29}]/2=>\sqrt{x}=[3+\sqrt{29}]/2<=>x=[19+3\sqrt{29}]/2` (t/m)
Vậy `S={[19+3\sqrt{29}]/2}`