1) TXĐ: $x>\frac{-1}{2}$<=> $\sqrt{x^2-\frac{1}{4}+\sqrt{(x+\frac{1}{2})^2}}=\frac{1}{2}(2x+1)(x^2+1)$<=> $\sqrt{x^2+x+\frac{1}{4}}=(x+\frac12)(x^2+1)$<=> $\sqrt{(x+\frac{1}{2})^2}=(x+\frac12)(x^2+1)$<=> $x+\frac12=(x+\frac12)(x^2+1)$<=> $x^2+1=1$<=> $x=0$ $(tm)$Vậy $S= $ {$0$}
1) TXĐ: $x
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eq \frac{-1}{2}$<=> $\sqrt{x^2-\frac{1}{4}+\sqrt{(x+\frac{1}{2})^2}}=\frac{1}{2}(2x+1)(x^2+1)$<=> $\sqrt{x^2+x+\frac{1}{4}}=(x+\frac12)(x^2+1)$<=> $\sqrt{(x+\frac{1}{2})^2}=(x+\frac12)(x^2+1)$<=> $x+\frac12=(x+\frac12)(x^2+1)$<=> $x^2+1=1$<=> $x=0$ $(tm)$Vậy $S= $ {$0$}