Đáp án + giải thích các bước giải:
$ĐKXD: \begin{cases} \dfrac{8x^3+1}{x+2}\ge0 \\ x+2\ge 0 \\4x^2-2x+1 \ge0 \\2x+1\ge0\end{cases} \\\to \begin{cases} \dfrac{(2x+1)(4x^2-2x+1)}{x+2} \ge0 \\x\ge-2 \\x\ge\dfrac{-1}{2} \end{cases} \\\to \begin{cases} \dfrac{2x+1}{x+2} \ge0 \\x\ge \dfrac{-1}{2} \end{cases} \\\to x\ge\dfrac{-1}{2}$
`\sqrt{(8x^3+1)/(x+2)}+\sqrt{x+2}=\sqrt{2x+1}+\sqrt{4x^2-2x+1}`
`->\sqrt{((2x+1)(4x^2-2x+1))/(x+2)}+\sqrt{x+2}=\sqrt{2x+1}+\sqrt{4x^2-2x+1}`
Đặt `a=\sqrt{2x+1}>=0;\sqrt{4x^2-2x+1}=b>0;\sqrt{x+2}=c>0`
`->(ab)/c+c=a+b`
`->ab+c^2=ac+bc`
`->c^2-ac+ab-bc=0`
`->c(c-a)+b(a-c)=0`
`->(c-b)(c-a)=0`
`->`\(\left[ \begin{array}{l}\sqrt{x+2}=\sqrt{4x^2-2x+1}\\\sqrt{x+2}=\sqrt{2x+1}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x+2=4x^2-2x+1\\x+2=2x+1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}4x^2-3x-1=0\\x=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}4x^2-4x+x-1=0\\x=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}(4x+1)(x-1)=0\\x=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{-1}{4}\\x=1\end{array} \right.(TM)\)
Vậy `S={-1/4;1}`
