Câu hỏi của hoàng thiên

Question

Chứng minh:

$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=1+\sqrt{3}$

♥ Aoko ♥ · Accepted Answer

Đặt $A=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}$

$A=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{4}.\sqrt{12}}}}$

$A=\sqrt{6+2\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}$

$A=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}$

$A=\sqrt{6+2\sqrt{5-\sqrt{12}-1}}$

$A=\sqrt{6+2\sqrt{4-\sqrt{12}}}$

$A=\sqrt{6+2\sqrt{4-\sqrt{4}.\sqrt{3}}}$

$A=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}$

$A=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}$

$A=\sqrt{6+2\left(\sqrt{3}-1\right)}$

$A=\sqrt{6+2\sqrt{3}-2}$

$A=\sqrt{4+2\sqrt{3}}$

$A=\sqrt{3+2\sqrt{3}+1}$

$A=\sqrt{\left(\sqrt{3}+1\right)^2}$

$A=1+\sqrt{3}$          (đpcm)

Vậy $A=1+\sqrt{3}$Câu trả lời: Đặt $A=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}$