Đáp án:
Giải thích các bước giải:
`x^2 + x + 6 + 2x\sqrt{x+3} = 4 ( x + \sqrt{x + 3 })`
ĐK: `x \ge -3`
`⇔ x^2-3x+2x\sqrt{x+3}-4\sqrt{x+3}+6=0`
`⇔ (x+\sqrt{x+3})^2-4x-4\sqrt{x+3}+3=0`
`⇔ (x+\sqrt{x+3})^2-4(x+\sqrt{x+3})+3=0`
Đặt `x+\sqrt{x+3}=t` ta có:
`t^2-4t+3=0`
`⇔ (t-1)(t-3)=0`
`⇔` \(\left[ \begin{array}{l}t=1\\t=3\end{array} \right.\)
`+) t=1⇒x+\sqrt{x+3}=1`
`⇔ \sqrt{x+3}=1-x`
ĐK: `x \le 1`
`⇔ x+3=1-2x+x^2`
`⇔ x^2-3x-2=0`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{3+\sqrt{17}}{2}\ (L)\\x=\dfrac{3-\sqrt{17}}{2}\ (TM)\end{array} \right.\)
`+) t=3⇒ x+\sqrt{x+3}=3`
`⇔ \sqrt{x+3}=3-x`
ĐK: `x \le 3
`⇔ x+3=9-6x+x^2`
`⇔ x^2-7x+6=0`
`⇔ (x-1)(x-6)=0`
`⇔` `⇔` \(\left[ \begin{array}{l}x=1\ (TM)\\x=6\ (L)\end{array} \right.\)
Vậy `S={\frac{3-\sqrt{17}}{2};1}`
