`\text{~~Holi~~}`
`a.(2x+1)(2-3x)=0`
`->`\(\left[ \begin{array}{l}2x+1=0\\2-3x=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}\end{array} \right.\)
Vậy `S={-1/2;2/3}`
`b. 2x(x+1)=x^2-1`
`-> 2x(x+1)=(x-1)(x+1)`
`-> 2x=(x-1)`
`-> 2x-(x-1)=0`
`-> 2x-x+1=0`
`-> x+1=0`
`-> x=-1`
Vậy `S={-1}`
`c. 3(x-2)+x^2+4=0`
`-> 3x-6+x^2+4=0`
`-> x^2+3x-2=0`
`-> (x+3/2)^2-17/4=0`
`-> (x+3/2)^2-(\sqrt{17}/2)^2=0`
`->(x+3/2-\sqrt{17}/2)(x+3/2+\sqrt{17}/2)=0`
`->`\(\left[ \begin{array}{l}x+\dfrac{3}{2}-\dfrac{\sqrt{17}}{2}=0\\x+\dfrac{3}{2}+\dfrac{\sqrt{17}}{2}=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-\dfrac{3-\sqrt{17}}{2}\\x=-\dfrac{3+\sqrt{17}}{2}\end{array} \right.\)
Vậy `S={-\frac{3±\sqrt{17}}{2}}`
`d. 3x^2+6x-9=0`
`-> 3(x^2+2x-3)=0`
`-> x^2+2x-3=0`
`-> x^2-x+3x-3=0`
`-> x(x-1)+3(x-1)=0`
`-> (x-1)(x+3)=0`
`->`\(\left[ \begin{array}{l}x-1=0\\x+3=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
Vậy `S={1;-3}`
