Câu hỏi của Nguyễn Thùy Chi

Question

Tính

A= $\sqrt{4+\sqrt{10+2\sqrt{5}}}$ + $\sqrt{4-\sqrt{10+2\sqrt{5}}}$

Nguyễn Việt Lâm · Accepted Answer

$A>0$

$A^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}$

$=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}$

$=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2$

$\Rightarrow A=\sqrt{5}+1$

Câu trả lời:

$A>0$

$A^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}$

$=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}$

$=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2$

$\Rightarrow A=\sqrt{5}+1$