Đáp án:
Giải thích các bước giải:
$(\sqrt{6+4\sqrt{2}})+(\sqrt{6-4\sqrt{2}})$
$=(\sqrt{(\sqrt{2})^2+2.\sqrt{2}.2+2^2})+(\sqrt{(\sqrt{2})^2-2.\sqrt{2}.2+2^2})$
$=\sqrt{(\sqrt{2}+2)^2}+\sqrt{(\sqrt{2}-2)^2}$
$=|\sqrt{2}+2|+|\sqrt{2}-2|$
$=(\sqrt{2}+2)+(2-\sqrt{2})$
$=\sqrt{2}+2+2-\sqrt{2}$
$=4$
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$\sqrt{x^2-2x+1}=(\sqrt{6+4\sqrt{2}})+(\sqrt{6-4\sqrt{2}})$
$⇔\sqrt{(x-1)^2}=4$
$⇔|x-1|=4$
$⇔\left[ \begin{array}{l}x-1=4\\x-1=-4\end{array} \right.$
$⇔\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.$
Vậy $x=5$ hoặc $x=-3$
