Câu hỏi của Mưa Đang Đi Chơi

Question

$\sqrt{\dfrac{2}{2-\sqrt{3}}}$

HT.Phong (9A5) · Accepted Answer

$\sqrt{\dfrac{2}{2-\sqrt{3}}}$$=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}$$=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{2^2-\left(\sqrt{3}\right)^2}}$$=\sqrt{\dfrac{4+2\sqrt{3}}{4-3}}$$=\sqrt{4+2\sqrt{3}}$$=\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}$$=\sqrt{\left(\sqrt{3}+1\right)^2}$$=\left|\sqrt{3}+1\right|$$=\sqrt{3}+1$Câu trả lời: $\sqrt{\dfrac{2}{2-\sqrt{3}}}$$=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}$$=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{2^2-\left(\sqrt{3}\right)^2}}$$=\sqrt{\dfrac{4+2\sqrt{3}}{4-3}}$$=\sqrt{4+2\sqrt{3}}$$=\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}$$=\sqrt{\left(\sqrt{3}+1\right)^2}$$=\left|\sqrt{3}+1\right|$$=\sqrt{3}+1$

Nguyễn Lê Phước Thịnh · Answer

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

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)