ĐK: `x>=-3`

`5x^2-4xsqrt(x+3)-x+4=0`

`<=>(5x^2-x+4)=4xsqrt(x+3)`

`<=>(5x^2-x+4)^2=(4xsqrt(x+3))^2`

`<=>[(5x^2-x)+4]^2=16x^2(x+3)`

`<=>(5x^2-x)^2+2.(5x^2-x).4+4^2=16x^3+48x^2`

`<=>25x^4-10x^3+x^2+40x^2-8x+16-16x^3-48x^2=0`

`<=>25x^4-26x^3-7x^2-8x+16=0`

`<=>25x^4-x^3-25x^3+x^2-8x^2-8x+16=0`

`<=>25x^4-25x^3-x^3+x^2-8x^2+8x-16x+16=0`

`<=>25x^3(x-1)-x^2(x-1)-8x(x-1)-16(x-1)=0`

`<=>(x-1)(25x^3-x^2-8x-16)=0`

Trường hợp `1: x-1=0<=>x=1(tm)`

Trường hợp `2:`

`25x^3-x^2-8x-16=0`

`<=>25x^3-25x^2+24x^2-24x+16x-16=0`

`<=>25x^2(x-1)+24x(x-1)+16(x-1)=0`

`<=>(x-1)(25x^2+24x+16)=0`

  `*` TH1: `x-1=0<=>x=1(tm)`

  `*` TH2: `25x^2+24x+16=0`

      Ta có: `a=25;b=24;c=16`

            `->Δ=b^2-4ac=24^2-4.2.16=-1024<0`

            `->` Phương trình vô nghiệm.

Vậy `S={1}`

`#owen`