Đáp án: `S={\frac{2±\sqrt{2}}{2};\frac{-3±\sqrt{7}}{2}}`
Giải thích các bước giải:
$(2x^2-3x+1)(2x^2+5x+1)=9x^2$
$⇔[(2x^2+x+1)-4x][(2x^2+x+1)+4x]=9x^2$
$⇔(2x^2+x+1)^2-16x^2=9x^2$
$⇔(2x^2+x+1)^2-25x^2=0$
$⇔(2x^2+x+1-5x)(2x^2+x+1+5x)=0$
$⇔(2x^2-4x+1)(2x^2+6x+1)=0$
$⇔\left[ \begin{array}{l}2x^2-4x+1=0(1)\\2x^2+6x+1=0(2)\end{array} \right.$
`(1)⇔x^2-2x+\frac{1}{2}=0⇔x^2-2x+1=\frac{1}{2}`
`⇔(x-1)^2=\frac{1}{2}⇔x-1=\frac{±\sqrt{2}}{2}⇔x=\frac{2±\sqrt{2}}{2}`
`(2)⇔x^2+3x+\frac{1}{2}=0⇔x^2+3x+\frac{9}{4}=\frac{7}{4}`
`⇔(x+\frac{3}{2})^2=\frac{7}{4}⇔x+\frac{3}{2}=\frac{±\sqrt{7}}{2}⇔x=\frac{-3±\sqrt{7}}{2}`
