$\sqrt[]{19+8\sqrt[]{3}}$ -$\sqrt[]{28-6\sqrt[]{3}}$+$\sqrt[]{12}$
=$\sqrt[]{3+2.4\sqrt[]{3}+16}$-$\sqrt[]{28-2\sqrt[]{27}}$+$\sqrt[]{4.3}$
=$\sqrt[]{({\sqrt[]{3}+4)^2}}$-$\sqrt[]{({3\sqrt[]{3}-1)^2}}$+2$\sqrt[]{3}$
=${\sqrt[]{3}+4}$-${3\sqrt[]{3}+1}$+2$\sqrt[]{3}$
`=5`
$\sqrt[]{x-3}$-2$\sqrt[]{x^2-9}$=0 (1)
ĐKXĐ: `x-3>=0; x^2-9>=0`
`<=>`\(\left[ \begin{array}{l}x≥3\\x≤-3\end{array} \right.\)
`(1)=>`$\sqrt[]{x-3}$-2$\sqrt[]{(x-3)(x+3)}$=0
`<=>`$\sqrt[]{x-3}$(1-2$\sqrt[]{x+3}$)=0
`<=>`\(\left[ \begin{array}{l}\sqrt[]{x-3}=0\\1-2\sqrt[]{x+3}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x-3=0\\2\sqrt[]{x+3}=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\\sqrt[]{x+3}=\frac{1}{2} \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x+3=\frac{1}{4}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=\frac{1}{4}-3 \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3(TM)\\x=\frac{-11}{4}(loại) \end{array} \right.\)
vậy `S={3}`
