Đáp án: $A=1$
Giải thích các bước giải:
Ta có:
$B=\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}$
$\to B^2=(\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}})^2$
$\to B^2=7+\sqrt{5}+7-\sqrt{5}+2\sqrt{7+\sqrt{5}}\sqrt{7-\sqrt{5}}$
$\to B^2=14+2\sqrt{7^2-5}$
$\to B^2=14+2\sqrt{44}$
$\to B^2=14+4\sqrt{11}$
$\to B^2=2(7+2\sqrt{11})$
$\to B=\sqrt{2}\cdot\sqrt{7+2\sqrt{11}}$
$\to \dfrac{B}{\sqrt{7+2\sqrt{11}}}=\sqrt{2}$
$\to \dfrac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}=\sqrt{2}$
$\to A=\sqrt{2}-\sqrt{3-2\sqrt{2}}$
$\to A=\sqrt{2}-\sqrt{2-2\sqrt{2}+1}$
$\to A=\sqrt{2}-\sqrt{(\sqrt{2}-1)^2}$
$\to A=\sqrt{2}-(\sqrt{2}-1)$
$\to A=1$
