Đáp án: $1+\sqrt{5}$
Giải thích các bước giải:
Đặt $x=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4+\sqrt{10-2\sqrt{5}}}(x>0)$
$⇒x^2=(4+\sqrt{10+2\sqrt{5}})+(4+\sqrt{10-2\sqrt{5}})+2.\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4+\sqrt{10-2\sqrt{5}}}$
$=8+2.\sqrt{(4+\sqrt{10+2\sqrt{5}})(4+\sqrt{10-2\sqrt{5}})}$
$=8+2.\sqrt{16-10-2\sqrt{5}}$
$=8+2.\sqrt{6-2\sqrt{5}}$
$=8+2\sqrt{5-2\sqrt{5}+1}$
$=8+2.\sqrt{(\sqrt{5}-1)^2}$
$=8+2.|\sqrt{5}-1|$
$=8+2(\sqrt{5}-1)$
$=8+2\sqrt{5}-2=6+2\sqrt{5}$
$=5+2\sqrt{5}+1$
$=(\sqrt{5}+1)^2$
$⇒x=|\sqrt{5}+1|=\sqrt{5}+1$
