Question

[lê thị bảo ngọc]

Bài 1: Tính

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

b> \(\sqrt{5+2\sqrt{6}}\) + \(\sqrt{5-2\sqrt{6}}\)

c> \(\sqrt{8-2\sqrt{7}}\) - \(\sqrt{8+2\sqrt{7}}\)

d> \(\sqrt{29+12\sqrt{5}}\) + \(\sqrt{29-12\sqrt{5}}\)

e> ( \(\sqrt{0,25}\) - \(\sqrt{225}\) + \(\sqrt{2,25}\)) : \(\sqrt{169}\)

f> 3 - \(\sqrt{5}\) + 3 + \(\sqrt{5}\)

Answer 1

[Trần Quốc Lộc]

\(\text{a) }\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\\ =\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}+1+\sqrt{5}-1\\ =2\sqrt{5}\)

\(\text{b) }\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\\ =\sqrt{3+2+2\sqrt{6}}+\sqrt{3+2-2\sqrt{6}}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)

\(\text{c) }\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\\ =\sqrt{7+1-2\sqrt{7}}-\sqrt{7+1+2\sqrt{7}}\\ =\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\\ =\sqrt{7}-1-\sqrt{7}-1\\ =-2\)

\(\text{d) }\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\\ =\sqrt{20+9+12\sqrt{5}}+\sqrt{20+9-12\sqrt{5}}\\ =\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\\ =\sqrt{20}+3+\sqrt{20}-3\\ =2\sqrt{20}\\ =4\sqrt{5}\)

\(\text{e) }\left(\sqrt{0,25}-\sqrt{225}+\sqrt{2,25}\right):\sqrt{169}\\ =\left(0,5-15+1,5\right):13\\ =\left(-13\right):13=-1\)

\(\text{f) }3-\sqrt{5}+3+\sqrt{5}\\ =6\)