Đáp án: `S={1; 2}`
Giải:
`\sqrt{5x^2+x+3}-2\sqrt{5x-1}+x^2-3x+3=0`
Điều kiện: `x≥\frac{1}{5}`
Pt ⇔ `\sqrt{5x^2+x+3}-(2x+1)-2[\sqrt{5x-1}-(x+1)]+x^2-3x+3+2x+1-2(x+1)=0`
⇔ `\frac{5x^2+x+3-(2x+1)^2}{\sqrt{5x^2+x+3}+2x+1}-\frac{2[5x-1-(x+1)^2]}{\sqrt{5x-1}+x+1}+x^2-3x+2=0`
⇔ `\frac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+\frac{2(x^2-3x+2)}{\sqrt{5x-1}+x+1}+x^2-3x+2=0`
⇔ `(x^2-3x+2)(\frac{1}{\sqrt{5x^2+x+3}+2x+1}+\frac{2}{\sqrt{5x-1}+x+1}+1)=0`
⇔ $\left [\begin{array}{l} x^2-3x+2=0 \\ \dfrac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+\dfrac{2}{\sqrt{5x-1}+x+1}+1=0 \end{array} \right.$
⇔ `(x-1)(x-2)=0`
Vì theo điều kiện thì `\frac{1}{\sqrt{5x^2+x+3}+2x+1}+\frac{2}{\sqrt{5x-1}+x+1}+1>0`
⇔ $\left [\begin{array}{l} x=1 & (tm) \\ x=2 & (tm) \end{array} \right.$
Vậy `S={1; 2}`
