Chứng minh rằng : \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
Bài 1: Chứng minh:
a) \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}=0\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
c) \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
d) \(\sqrt{6+\sqrt{6+\sqrt{6+,,,+\sqrt{6}}}}=3\)
Chứng minh:
\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{30}+\sqrt{42}
\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3\sqrt{\left(\sqrt{20-3}\right)^2}}\)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Cho mình hỏi:
a.\(\sqrt{15-6\sqrt{6}}+\sqrt{42-12\sqrt{6}}\)
b.1\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{24}+\sqrt{25}}\)
Trả lời giúp mình với ạ! mình cảm ơn!
Rút gọn biểu thức:
\(a,\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(b,\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
Câu 1: Tính
a. \(\sqrt{3\dfrac{6}{25}}\) b. \(\sqrt[3]{261}\) c. \(\sqrt{8,1}\) . \(\sqrt{20}\). \(\sqrt{8}\)
d. \(\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}\)
\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(\sqrt{30+12\sqrt{6}}+\)\(\sqrt{30-12\sqrt{6}}\)
\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)\(-\sqrt{2}\)
