Bài 1:
`a) 3\sqrt{x^2-1}-\sqrt{x-1}=0` ĐK: `x>=1`
`<=> 3\sqrt{x-1}. \sqrt{x+1}-\sqrt{x-1}=0`
`<=> \sqrt{x-1}(3\sqrt{x+1}-1)=0`
`<=>`\(\left[ \begin{array}{l}\sqrt{x-1}=0\\3\sqrt{x+1}=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x-1=0\\\sqrt{x+1}=\dfrac{1}{3}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x+1=\dfrac{1}{9}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1(tm)\\x=\dfrac{-8}{9} (ktm)\end{array} \right.\)
Vậy `S={1}`
`b) \sqrt{2x-3}-\sqrt{4x^2-9}=0` ĐK: `x>=3/2`
`<=> \sqrt{2x-3}-\sqrt{2x-3}.\sqrt{2x+3}=0`
`<=> \sqrt{2x-3}(1-\sqrt{2x+3})=0`
`<=>`\(\left[ \begin{array}{l}\sqrt{2x-3}=0\\\sqrt{2x+3}=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x-3=0\\2x+3=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{3}{2}(tm)\\x=-1(ktm)\end{array} \right.\)
Vậy `S={3/2}`
