Đáp án:
`x=\frac{\sqrt{97}+9}{2}` hoặc `x=\frac{-\sqrt{97}+9}{2}`
Giải thích các bước giải:
`-6x-3=3x-x^2+1`
`<=>x^2-3x-6x-3-1=0`
`<=>x^2-9x-4=0`
`<=>x^2-9/2x-9/2x+81/4-97/4=0`
`<=>x(x-9/2)-9/2(x-9/2)=97/4`
`<=>(x-9/2)(x-9/2)=97/4`
`<=>(x-9/2)^2=97/4`
\(⇔\left[ \begin{array}{l}x-\dfrac92=\dfrac{\sqrt{97}}{2}\\x-\dfrac92=\dfrac{-\sqrt{97}}{2}\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{\sqrt{97}+9}{2}\\x=\dfrac{-\sqrt{97}+9}{2}\end{array} \right.\)
Vậy `x=\frac{\sqrt{97}+9}{2}` hoặc `x=\frac{-\sqrt{97}+9}{2}`
