Đáp án:
`S={(3+3\sqrt5)/2;(3-3\sqrt5)/2}`
Giải thích các bước giải:
` x^2+(9x^2)/(x+3)^2=27`
`<=>x^2-2x . (3x)/(x+3)+(3x)^2/(x+3)^2+(6x^2)/(x+3)-27=0`
`<=>(x-(3x)/(x+3))^2+(6x^2)/(x+3)-27=0`
`<=>((x^2-3x+3x)/(x+3))^2+(6x^2)/(x+3)-27=0`
`<=>((x^2)/(x+3))^2+(6x^2)/(x+3)-27=0`
Đặt `(x^2)/(x+3)=t`, ta có:
`t^2+6t-27=0`
`<=>(t-3)(t+9)=0<=>` $\left[ \begin{array}{l}t=3\\t=-9\end{array} \right.$
+) Nếu` t=3=>(x^2)/(x+3)=3`
`<=>x^2=3(x+3)`
`<=>x^2-3x-9=0`
`<=>`\(\left[ \begin{array}{l}x=\dfrac{3+3\sqrt5}{2}\\x=\dfrac{3-3\sqrt5}{2}\end{array} \right.\)
+)Nếu` t=-9,` ta có:
`(x^2)/(x+3)=-9`
`<=>x^2=-9x-27`
`<=>x^2+9x+27=0`
`<=>x^2+2.x . 9/2+9/4+99/4=0`
`<=>(x+9/2)^2+99/4>=99/4>0`
`=>` pt VN
Vậy`S={(3+3\sqrt5)/2;(3-3\sqrt5)/2}`
