Đáp án + Giải thích các bước giải:
`x + 16 - 6\sqrt{2x + 1} = 2\sqrt{5 - x}` `(ĐK : -1/2 ≤ x ≤ 5)`
`⇔ x+16-6\sqrt{2x + 1}-2\sqrt{5 - x}=0`
`⇔ (2x + 1 - 6\sqrt{2x + 1} + 9) + (5 - x - 2\sqrt{5 - x} + 1) = 0`
`⇔[\sqrt{(2x+1)^2}-3.\sqrt{2x+1}.3 + 3^2]+[\sqrt{(5-x)^2}-2\sqrt{5-x}.1 + 1^2]=0`
`⇔ (\sqrt{2x + 1} - 3)^2 + (\sqrt{5 - x} - 1)^2 = 0`
$⇔\left[ \begin{array}{l}\sqrt{2x+1}-3=0\\\sqrt{5-x}-1=0\end{array} \right.$
$⇔\left[ \begin{array}{l}\sqrt{2x+1}=3\\\sqrt{5-x}=1\end{array} \right.$
$⇔\left[ \begin{array}{l}2x+1=9\\5-x=1\end{array} \right.$
$⇔\left[ \begin{array}{l}x=4(N)\\x=4(N)\end{array} \right.$
Vậy `S={4}`
