Question

[Lê Hồng Ngọc]

tính B=\(\sqrt{3\cdot\sqrt{5}}\cdot\left(3+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)

Answer 1

[Nguyễn Đức Tuệ]

Đây phải là dấu cộng

B=\(\sqrt{3+\sqrt{5}}.\left(3+\sqrt{5}\right).\sqrt{2}.\left(\sqrt{5}-1\right)\)

B=\(\sqrt{6+2\sqrt{5}}.\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)\)

B=\(\sqrt{\left(\sqrt{5}+1\right)^2}.\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)\)

B=\(\left(\sqrt{5}+1\right).\left(\sqrt{5}-1\right).\left(3+\sqrt{5}\right)\)

B=12+4\(\sqrt{5}\)