Đáp án:
`S={2+-sqrt5}`
Giải thích các bước giải:
Điều kiện: `x>=-1/3`
`(x+1)sqrt(8x+5)+sqrt(6x+2)=x^2+4x+3`
`<=>x^2+4x+3-(x+1)sqrt(8x+5)-sqrt(6x+2)=0`
`<=>x^2+3x+2-(x+1)sqrt(8x+5)+x+1-sqrt(6x+2)=0`
`<=>(x+1)(x+2)-(x+1)sqrt(8x+5)+x+1-sqrt(6x+2)=0`
`<=>(x+1)(x+2-sqrt(8x+5))+(x+1-sqrt(6x+2))=0`
`<=>(x+1)((x+2-sqrt(8x+5))(x+2+sqrt(8x+5)))/(x+2+sqrt(8x+5))+((x+1-sqrt(6x+2))(x+2+sqrt(6x+2)))/(x+1+sqrt(6x+2))=0`
`<=>(x+1)(x^2-4x-1)/(x+2+sqrt(8x+5))+(x^2-4x-1)/(x+1+sqrt(6x+2))=0`
`<=>(x^2-4x-1)((x+1)/(x+2+sqrt(8x+5))+1/(x+1+sqrt(6x+2)))=0`
Với `x>=-1/3=>(x+1)/(x+2+sqrt(8x+5))+1/(x+1+sqrt(6x+2))>0`
`=>x^2-4x-1=0`
`<=>x^2-4x+4-5=0`
`<=>(x-2)^2=5`
`<=>`\(\left[\begin{array}{l}x-2=\sqrt{5}\\x-2=-\sqrt{5}\end{array}\right.\)
`<=>`\(\left[\begin{array}{l}x=\sqrt{5}+2\\x=-\sqrt{5}+2\end{array}\right.\)
Vậy pt có tập nghiệm `S={2+-sqrt5}`
