Câu hỏi của Bắp Ngô

Question

Tính:$\sqrt{3+\sqrt{3}}+\sqrt{2-\sqrt{3}}$

l҉o҉n҉g҉ d҉z҉ · Accepted Answer

gọi $A=\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}$

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

$< =>A^2=6+2\sqrt{9-3\sqrt{3}+3\sqrt{3}-\sqrt{3^2}}$

$< =>A^2=6+2\sqrt{6}$

$< =>A=\sqrt{6+2\sqrt{6}}$

Câu trả lời:

gọi $A=\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}$

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

$< =>A^2=6+2\sqrt{9-3\sqrt{3}+3\sqrt{3}-\sqrt{3^2}}$

$< =>A^2=6+2\sqrt{6}$

$< =>A=\sqrt{6+2\sqrt{6}}$