Đáp án:
`4x+\sqrt{3-2x^2}=x^2+4`(ĐKXĐ: `-\sqrt{6}/2\le x \le\sqrt{6}/2`)
`<=> \sqrt{3-2x^2}=x^2+4-4x`
`<=> 3-2x^2=x^4+16+16x^2+8x^2-8x^3-32x`
`<=> 3-2x^2=x^4+16+24x^2-8x^2-32x`
`<=> 3-2x^2-x^4-16-24x^2+8x^2+32x=0`
`<=> -13-26x^2-x^4+8x^3+32x=0`
`<=> -x^4+8x^3-26x^2+32x-13=0`
`<=> -x^4+x^3+7x^3-7x^2-19x^2+19x+13x-13=0`
`<=> -(x-1)(x^3-7x^2+19x-13)=0`
`<=> -(x-1)(x-1)(x^2-6x+13)=0`
`<=> -(x-1)^2(x^2-6x+13)=0`
`<=>`\(\left[ \begin{array}{l}(x-1)^2=0\\x^2-6x+13=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x\notin\mathbb{R}\end{array} \right.\)
`<=> x=1`(tm)
Vậy `S={1}`
