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