Question

[Cá Lệ Kiều]

a) \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

b) \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

c) \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

đề bài là rút gọn biểu thức

giải chi tiết hộ mình ạ !!!

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)

\(=\sqrt{3}-1\)

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

\(=3-2\sqrt{2}+3\sqrt{2}+1\)

\(=4+\sqrt{2}\)

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

\(=2\sqrt{2}-2+2\sqrt{2}+1\)

\(=4\sqrt{2}-1\)

Answer 2

[Nguyen Minh Hieu]

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)