Đáp án + giải thích các bước giải:
`\sqrt{x^2-1/4+\sqrt{x^2+x+1/4}}=1/2(2x^3+x^2+2x+1)`
`->\sqrt{x^2-1/4+\sqrt{(x+1/2)^2}}=1/2[2x^2(x+1/2)+2(x+1/2)]`
`->\sqrt{x^2-1/4+\sqrt{(x+1/2)^2}}=1/2(2x^2+2)(x+1/2)`
`->\sqrt{x^2-1/4+|x+1/2|}=(x^2+1)(x+1/2) `
Điều kiện xác định: $\left\{\begin{matrix} x^2-\dfrac{1}{4}+|x+\dfrac{1}{2}|\ge0\\(x^2+1)(x+\dfrac{1}{2})\ge0 \end{matrix}\right.\\\to\left\{\begin{matrix} x^2-\dfrac{1}{4}+|x+\dfrac{1}{2}|\ge0\\x+\dfrac{1}{2}\ge0 \end{matrix}\right. \\\to\left\{\begin{matrix} x^2-\dfrac{1}{4}+|x+\dfrac{1}{2}|\ge0\\x\ge \dfrac{-1}{2} \end{matrix}\right. \\\to\left\{\begin{matrix} x^2-\dfrac{1}{4}+x+\dfrac{1}{2}\ge0\\x\ge\dfrac{-1}{2} \end{matrix}\right.\\\to x\ge \dfrac{-1}{2}$
`->\sqrt{x^2-1/4+x+1/2}=(x^2+1)(x+1/2)`
`->\sqrt{x^2+x+1/4}=(x^2+1)(x+1/2)`
`->\sqrt{(x+1/2)^2}=(x^2+1)(x+1/2)`
`->|x+1/2|=(x^2+1)(x+1/2)`
`->x+1/2=(x^2+1)(x+1/2)`
`->(x+1/2)(x^2+1-1)=0`
`->(x+1/2)x^2=0`
`->`\(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=0\end{array} (TM)\right.\)
