Đáp án :
`x=1`
Giải thích các bước giải :
`|x^2+|6x-2||=x^2+4``->`\(\left[ \begin{array}{l}x^2+|6x-2|=x^2+4\\x^2+|6x-2|=-x^2-4\end{array} \right.\)`->`\(\left[ \begin{array}{l}|6x-2|=4\\|6x-2|=-2x^2-4 (L)\end{array} \right.\)`->|6x-2|=4``->`\(\left[ \begin{array}{l}6x-2=4\\6x-2=-4\end{array} \right.\)`->`\(\left[ \begin{array}{l}6x=6\\6x=-2\end{array} \right.\)`->`\(\left[ \begin{array}{l}x=1 (Tm)\\x=\dfrac{-1}3 (L)\end{array} \right.\)Vậy : `x=1`
Giải:
| x² + | 6x - 2|| = x² + 4
Vì x² ≥ 0 nên x² + 4 ≥ 4 > 0
Và x² + | 6x - 2|| > 0 nên
x² + | 6x - 2 | = x² + 4
⇔ | 6x - 2 | = 4
⇔ 6x - 2 = 4 or 6x - 2 = -4
⇔ x = 1 or x = -1/3 ( loại vì x không nguyên )
Vậy S={ 1 }
@tienhcb123456789
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