`***`Lời giải`***`
`\sqrt{4x^2-9}=2\sqrt{2x+3}`
ĐK: `x≥3/2`
`<=>\sqrt{(2x-3)(2x+3)}=2\sqrt{2x+3}`
`<=>\sqrt{(2x-3)(2x+3)}-2\sqrt{2x+3}=0`
`<=>\sqrt{(2x+3)}(\sqrt{(2x-3)}-2)=0`
`<=>`\(\left[ \begin{array}{l}\sqrt{2x+3}=0\\\sqrt{2x-3}-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x+3=0\\\sqrt{2x-3}=2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x+3=0\\2x-3=4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=-3\\2x=7\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{-3}{2}(L)\\x=\dfrac{7}{2}(N)\end{array} \right.\)
Vậy `S={7/2}`
`\sqrt{x^2+2x+1}=\sqrt{x+1}`
ĐK: `x≥-1`
Bình phương 2 vế ta đươc:
`x^2+2x+1=x+1`
`<=>x^2+2x-x=1-1`
`<=>x^2+x=0`
`<=>x(x+1)=0`
`+)x=0(N)`
`+)x+1=0<=>x=-1(N)`
Vậy `S={0;-1}`
