Câu hỏi của Trần Hạnh Nguyên

Question

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

Mysterious Person · Accepted Answer

$\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}-\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}$

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

= $2\sqrt{2}-1$

Câu trả lời: $\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}-\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}$

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

= $2\sqrt{2}-1$